Class 11

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.2 Questions and Answers.

11th Maths Part 1 Trigonometry – II Exercise 3.2 Questions And Answers Maharashtra Board

Question 1.
Find the values of:
i. sin 690°
ii. sin 495°
iii. cos 315°
iv. cos 600°
v. tan 225°
vi. tan (- 690°)
vii. sec 240°
viii. sec (- 855°)
ix. cosec 780°
x. cot (-1110°)
Solution:
i. sin 690° = sin (720° -30°)
Solution:
i. sin 690° = sin (720° -30°)
= sin (2 x 360° – 30°)
= – sin 30°
= \(\frac{-1}{2}\)

ii. sin 495° = sin (360° + 135°)
= sin (135°)
= sin (90° + 45°)
= cos 45°
= \(\frac{1}{\sqrt{2}}\)

iii. cos 315° = cos (270° + 45°)
sin 45° = \(\frac{1}{\sqrt{2}}\)

iv. cos 600° = cos (360° + 240°)
= cos 240°
= cos (180° + 60°)
= – cos 60°
= \(-\frac{1}{2}\)

v. tan 225° = tan (180° + 45°)
= tan 45°
= 1 .

vi. tan (- 690°) = – tan 690°
= – tan (720° – 30°)
= – tan (2 x 360° – 30°)
= – (- tan 30°)
= tan 30°
= \(\frac{1}{\sqrt{3}}\)

vii. sec 240° = sec (180° + 60°)
= – sec 60°
= – 2

viii. sec (-855°) = sec (855°)
= sec (720°+135°)
= sec (2 x360°+ 135°) = sec 135°
= sec (90° + 45°)
= – cosec 45°
= –\(\sqrt{2}\)

ix. cosec 780° = cosec (720° + 60°)
= cosec (2 x 360° + 60°)
= cosec 60°
= \(\frac{2}{\sqrt{3}}\)

x. cot (-1110°) =-cot (1110°)
= -cot (1080°+ 30°)
= – cot (3 x 360° + 30° )
= – cot 30°
= – \(\sqrt{3}\)

Question 2.
Prove the following:
i. \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x\)
ii. \(\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]\)
iii. sec 840° cot (- 945°) + sin 600° tan (- 690°) = 3/2
iv. \(\frac{{cosec}\left(90^{\circ}-x\right) \sin \left(180^{\circ}-x\right) \cot \left(360^{\circ}-x\right)}{\sec \left(180^{\circ}+x\right) \tan \left(90^{\circ}+x\right) \sin (-x)}=1\)
v. \(\frac{\sin ^{3}(\pi+x) \sec ^{2}(\pi-x) \tan (2 \pi-x)}{\cos ^{2}\left(\frac{\pi}{2}+x\right) \sin (\pi-x) {cosec}^{2}(-x)}=\tan ^{3} x\)
vi. cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ) = 0
Solution:
i.

ii. L.H.S.
= cos ( \(\frac{3 \pi}{2}\) + x) cos (2π + x) . [cot ( – x) + (2π + x)]
= (sin x)(cos x) (tan x + cot x)
= sin x cos x ( \(\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\))
= sin x cos x \(\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)\)
= sin x cos x \(\left(\frac{1}{\sin x \cos x}\right)\)
= 1 = R.H.S

iii. sec 840° = sec (720° + 120°)
= sec (2 x 360° + 120°)
= sec (120°)
= sec (90° + 30°)
= – cosec 30°
= -2

cot(-945°) = -cot 945°
= -cot (720° + 225°)
= -cot (2 x 360° +225°)
= -cot (225°)
= -cot (180° + 459)
= -cot 45°
= -1

sin 600° = sin (360° + 240°)
= sin (240°)
= sin (180° +60°)
= – sin 60° = –\(\frac{\sqrt{3}}{2}\)

tan (-690°) = – tan 690°
= – tan (360° +330°)
= -tan (330°)
=- tan (360° – 30°)
=-(-tan 30°)
= tan 30°0 = \(\frac{1}{\sqrt{3}}\)

L.H.S. = sec 840° cot (-945°) + sin 600° tan (-690°)
= (-2)(-1) + \(\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{3}}\right)\)
= 2 – \(\frac{1}{2}=\frac{3}{2}\)
= R. H. S.

iv.

= 1
= R.H.S

v.

vi. L.H.S. = cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ)
= cos θ + (- cos θ)-(- cos θ) – cos θ
= cos θ – cos θ + cos θ – cos θ
= 0
= R.H.S.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 1
• Exercise 3.2 Class 11 Maths 1
• Exercise 3.3 Class 11 Maths 1
• Exercise 3.4 Class 11 Maths 1
• Exercise 3.5 Class 11 Maths 1
• Miscellaneous Exercise 3 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.1 Questions and Answers.

11th Maths Part 1 Trigonometry – II Exercise 3.1 Questions And Answers Maharashtra Board

Question 1.
Find the values of:
i. sin 150°
ü. cos 75°
iii. tan 105°
iv. cot 225°
Solution:
i. sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
\(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
[Note: Answer given in the textbook is \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\) However, as per our calculation it is \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

ii. cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°

iii. tan 105° = tan (60° +45°)

iv. cot 225°

Question 2.
Perove the following:
i. \(\cos \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-y\right) -\sin \left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-y\right)=-\cos (x+y)\)
Solution:
L.H.S

= -(cos x cos y – sin x sin y)
= – cos (x+y)
= R.H.S

ii. \(\tan \left(\frac{\pi}{4}+\theta\right)=\frac{1+\tan \theta}{1-\tan \theta}\)
L.H.S =\(\tan \left(\frac{\pi}{4}+\theta\right)\)

R.H.S.
[Note : The question has been modified.]

iii. \(\left(\frac{1+\tan x}{1-\tan x}\right)^{2}=\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}\)
Solution:

iv. sin [(n+1)A] . sin [(n+2)A] + cos [(n+1)A] . cos [(n+2)A] = cos A
Solution:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let(n+2)Aaand(n+l)Ab …(i)
∴ L.H.S. = cos a. cos b + sin a. sin b
= cos (a — b)
= cos [(n + 2)A — (n + I )A]
…[From (i)]
cos[(n+2 – n – 1)A]
= cos A
= R.H.S.

v. \(\sqrt{2} \cos \left(\frac{\pi}{4}-\mathrm{A}\right)=\cos \mathrm{A}+\sin \mathrm{A}\)
Solution:

vi. \(\frac{\cos (x-y)}{\cos (x+y)}=\frac{\cot x \cot y+1}{\cot x \cot y-1}\)
Solution:

vii. cos (x + y). cos (x – y) = cos²y – sin²x
Solution:
L.H.S. = cos(x + y). cos(x – y)
= (cos x cos y – sin x sin y). (cos x cos y + sin x sin y)
= cos² x cos²y – sin²x sin²y
…[∵ (a – b) (a + b) = a² – b²]
= (1 – sin²x) cos²y – sin²x (1 – cos²y)
…[∵ sin²e + cos²0 = 1]
= cos²y – cos²y sin²x – sin²x + sin²x cos²y
= cos²y – sin²x
=R.H.S.

viii.\(\frac{\tan 5 A-\tan 3 A}{\tan 5 A+\tan 3 A}=\frac{\sin 2 A}{\sin 8 A}\)
Solution:

ix. tan 8θ – tan 5θ – tan 3θ = tan 8θ tan 5θ tan 3θ
Solution:
Since, 8θ = 5θ + 3θ
∴ tan 8θ = tan (5θ + 3θ)
∴ tan 8θ = \(\frac{\tan 5 \theta+\tan 3 \theta}{1-\tan 5 \theta \tan 3 \theta}\)
∴ tan 8θ (1 – tan 5θ.tan 3θ) = tan 5θ + tan 3θ
∴ tan 8θ – tan8θ.tan5θ.tan3θ = tan5θ + tan 3θ
∴ tan 8θ – tan 5θ – tan 3θ = tan 8θ.tan 5θ.tan 3θ

x. tan 50° = tan 40° + 2tan 10°
Solution:
Since, 50° = 10° +40°
∴ tan 50° = tan (10° + 40°)
∴ \(\frac{\tan 10^{\circ}+\tan 40^{\circ}}{1-\tan 10^{\circ} \tan 40^{\circ}}\)
∴ tan 50° (1 – tan 10° tan 40°) = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° tan 50° = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° tan (90° – 40°) = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° cot 40°
= tan 10° + tan 40° …[∵ tan (90° – θ) = cot θ]
∴ tan 50° – tan 10° tan 40°. \(\frac{1}{\tan 40^{\circ}}\) = tan 10° + tan 40°
∴ tan 50° – tan 10°. 1 = tan 10° + tan 40°
∴ tan 50° = tan 40° + 2 tan 10°

xi. \(\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}\) = tan 72°
Solution:
\(\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}\)
Dividing numerator and cos 27°, we get denominator by cos 27°, we get

= tan (45° + 27°)
= tan 72° = R.H.S

xii. \(\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}=\tan 72^{\circ}\)
Solution:
Since 45° = 10° + 35°,
tan 45° = tan (10° +35°)
∴ \(\frac{\tan 10^{\circ}+\tan 35^{\circ}}{1-\tan 10^{\circ} \tan 35^{\circ}}\)
∴ 1 – tan 10° tan 35o = tan 10° + tan 35°
∴ tan 10° + tan 35° + tan 10° tan 35° = 1

xiii. tan 10° + tan 35° + tan 10°. tan 35° = 1
Solution:

xiv. \(\frac{\cos 15^{\circ}-\sin 15^{\circ}}{\cos 15^{\circ}+\sin 15^{\circ}}=\frac{1}{\sqrt{3}}\)
Solution:
Dividing numerator and cos 15°, we get

= tan (45° + 15°)
= tan 30° = \(\frac{1}{\sqrt{3}}\) = R.H.S

Question 3.
If sin A = \(-\frac{5}{13}\),π < A < \(\frac{3 \pi}{2}\) and cos B = \(\frac{3}{5}, \frac{3 \pi}{2}\) < B < 2π, find
i. sin (A+B)
ii. cos (A-B)
iii. tan (A + B)
Solution:
Given, sin A = \(-\frac{5}{13}\)
We know that,
cos² A = 1 – sin²A = \(1-\left(-\frac{5}{13}\right)^{2}=1-\frac{25}{169}=\frac{144}{169}\)
∴ cos A = \(\pm \frac{12}{13}\)
Since, π < A < \(\frac{3 \pi}{2}\)
∴ ‘A’ lies in the 3rd quadrant.
∴ cos A<0
cos A = \(\frac{-12}{13}\)
Also,cos B = \(\frac{3}{5}\)
∴ sin²B = 1 – cos²B = \(1-\left(\frac{3}{5}\right)^{2}=1-\frac{9}{25}=\frac{16}{25}\)
∴ sin B = \(\pm \frac{4}{5}\)
Since, \(\frac{3 \pi}{2}\) < B < 2π
∴ ‘B’ lies in the 4th quadrant.
∴ sin B<0
Sin B = \(\frac{-4}{5}\)

i. sin (A + B) = sin A cos B+cos A sin B

ii. cos (A -B) = cos A cos B + sin A sin B

iii.

Question 4.
If tan A = \(\frac{5}{6}\) , tan B = \(\frac{1}{11}\) prove that A + B = \(\frac{\pi}{4}\)
Solution:
Given tan A = \(\frac{5}{6}\), tan B = \(\frac{1}{11}\)

∴ tan (A + B) = tan \(\frac{\pi}{4}\)
∴ A + B = \(\frac{\pi}{4}\)

Class 11 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 11 Maths 1
• Exercise 3.2 Class 11 Maths 1
• Exercise 3.3 Class 11 Maths 1
• Exercise 3.4 Class 11 Maths 1
• Exercise 3.5 Class 11 Maths 1
• Miscellaneous Exercise 3 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Miscellaneous Exercise 2 Questions and Answers.

11th Maths Part 1 Trigonometry – I Miscellaneous Exercise 2 Questions And Answers Maharashtra Board

I. Select the correct option from the given alternatives.

Question 1.
The value of the expression
cos1°. cos2°. cos3° … cos 179° =
(A) -1
(B) 0
(C) \(\frac{1}{\sqrt{2}}\)
(D) 1
Answer:
(B) 0

Explanation:
cos 1° cos 2° cos 3° … cos 179°
= cos 1° cos 2° cos 3° … cos 90°… cos 179°
= 0 …[∵ cos 90° = 0]

Question 2.
\(\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}\) is equal to
(A) 2cosec A
(B) 2 sec A
(C) 2 sin A
(D) 2 cos A
Answer:
(A) 2cosec A

Explanation:

Question 3.
If α is a root of 25cos² θ + 5cos θ – 12 = 0, \(\frac{\pi}{2}\) < α < π, then sin 2α is equal to
(A) \(-\frac{24}{25}\)
(B) \(-\frac{13}{18}\)
(C) \(\frac{13}{18}\)
(D) \(\frac{24}{25}\)
Answer:
(A) \(-\frac{24}{25}\)

Explanation:

25 cos² θ + 5 cos θ – 12 = 0
∴ (5cos θ + 4) (5 cos θ – 3) = 0
∴ cos θ = \(-\frac{4}{5}\) or cos θ = \(\frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π,
cos α < 0
∴ cos α = \(-\frac{4}{5}\)
sin² α = 1 – cos² α = 1 – \(\frac{16}{25}=\frac{9}{25}\)
∴ sin α = \(\pm \frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π sin α > 0
∴ sin α = 3/5
sin 2 α = 2 sin α cos α
= \(2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}\)

Question 4.
If θ = 60°, then \(\frac{1+\tan ^{2} \theta}{2 \tan \theta}\) is equal to
(A) \(\frac{\sqrt{3}}{2}\)
(B) \(\frac{2}{\sqrt{3}}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt{3}\)
Answer:
(B) \(\frac{2}{\sqrt{3}}\)

Explanation:

Question 5.
If sec θ = m and tan θ = n, then \(\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}\) is equal to
(A) 2
(B) mn
(C) 2m
(D) 2n
Answer:
(A) 2
Explanation:

Question 6.
If cosec θ + cot θ = \(\frac{5}{2}\), then the value of tan θ is
(A) \(\frac{14}{25}\)
(B) \(\frac{20}{21}\)
(C) \(\frac{21}{20}\)
(D) \(\frac{15}{16}\)
Answer:
(B) \(\frac{20}{21}\)

Explanation:
cosec θ + cot θ = \(\frac{5}{2}\) …………….(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ \(\frac{5}{2}\) (cosec θ – cot θ) = 1
∴ cosec θ – cot θ = \(\frac{2}{5}\) …(ii)
Subtracting (ii) from (i), we get
2 cot θ = \(\frac{5}{2}-\frac{2}{5}=\frac{21}{10}\)
∴ cot θ = \(\frac{21}{20}\)
∴ tan θ = \(\frac{20}{21}\)

Question 7.
\(1-\frac{\sin ^{2} \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}\) equals
(A) 0
(B) 1
(C) sin θ
(D) cos θ
Answer:
(D) cos θ

Explanation:

Question 8.
If cosec θ – cot θ = q, then the value of cot θ is
(A) \(\frac{2 q}{1+q^{2}}\)
(B) \(\frac{2 q}{1-q^{2}}\)
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)
(D) \(\frac{1+q^{2}}{2 q}\)
Answer:
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

Explanation:

cosec θ – cot θ = q ……(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ (cosec θ + cot θ)q = 1
∴ cosec θ + cot θ = 1/q …….(ii)
Subtracting (i) from (ii), we get
2cot θ = \(\frac{1}{\mathrm{q}}-\mathrm{q}\)
∴ cot θ = \(\frac{1-q^{2}}{2 q}\)

Question 9.
The cotangent of the angles \(\frac{\pi}{3}, \frac{\pi}{4}\) and \(\frac{\pi}{6}\) are in
(A) A.P.
(B) G.P.
(C) H.P.
(D) Not in progression
Answer:
(B) G.P.

Explanation:

Question 10.
The value of tan 1°.tan 2° tan 3° equal to
(A) -1
(B) 1
(C) \(\frac{\pi}{2}\)
(D) 2
Answer:
(B) 1

Explanation:

tan1° tan2° tan3° … tan89°
= (tan 1° tan 89°) (tan 2° tan 88°)
…(tan 44° tan 46°) tan 45°
= (tan 1 ° cot 1 °) (tan 2° cot 2°)
…(tan 44° cot 44°) . tan 45°
…tan(∵ 90° – θ) = cot θ]
= 1 x 1 x 1 x … x 1 x tan 45° =1

II. Answer the following:

Question 1.
Find the trigonometric functions of:
90°, 120°, 225°, 240°, 270°, 315°, -120°, -150°, -180°, -210°, -300°, -330°
Solution:
Angle of measure 90° :
Let m∠XOA = 90°
Its terminal arm (ray OA)
intersects the standard, unit circle at P(0, 1).

∴ x = 0 and y = 1
sin 90° = y = 1
cos 90° = x = 0
tan 90° = \(\frac{y}{x}=\frac{1}{0}\), which is not defined
cosec 90° = \(\frac{1}{y}=\frac{1}{1}\) = 1
sec 90° = \(\frac{1}{x}=\frac{1}{0}\), which is not defined
cot 90° = \(\frac{x}{y}=\frac{0}{1}\) = 0

Angle of measure 120° :
Let m∠XOA =120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 2nd quadrant, x < 0, y > 0

[Note: Answer given in the textbook of tan 120° is \(\frac{-1}{\sqrt{3}}\) and cot 120° is \(-\sqrt{3}\). However, as per our \(-\sqrt{3}\) calculation the answer of tan 120° is \(-\sqrt{3}\) and cot 120° is \(-\frac{1}{\sqrt{3}}\)

Angle of measure 225° :
Let m∠XOA = 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 240° :
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 270° :
Let m∠XOA = 270°
Its terminal arm (ray OA) intersects the standard unit circle at P(0, – 1).
x = 0 andy = – 1
sin 270° = y = -1
cos 270° = x = 0
tan 270° = \(\frac{y}{x}\)

Angle of measure 315° :
Let m∠XOA = 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1


[Note: Answer given in the textbook of cot 315° is 1. However, as per our calculation it is -1.]

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (-150°) :
Let m∠XOA = – 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (-180°):
Let m∠XOA = – 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(- 1, 0).
∴ x = – 1 andy = 0
sin (-180°) = y = 0
cos (-180°) = x
= -1

Angle of measure (- 210°):
Let m∠XOA = -210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (- 300°):
Let m∠XOA = – 300° Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 1st quadrant, x>0,y>0
x = OM = \(\frac{1}{2}\) and
y = PM = \(\frac{\sqrt{3}}{2}\)

Angle of measure (- 330°):
Let m∠XOA = – 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Question 2.
State the signs of:
i. cosec 520°
ii. cot 1899°
iii. sin 986°
Solution:
i. 520° =360° + 160°
∴ 520° and 160° are co-terminal angles.
Since 90° < 160° < 180°,
160° lies in the 2nd quadrant.
∴ 520° lies in the 2nd quadrant,
∴ cosec 520° is positive.

ii. 1899° = 5 x 360° + 99°
∴ 1899° and 99° are co-terminal angles.
Since 90° < 99° < 180°,
99° lies in the 2nd quadrant.
∴ 1899° lies in the 2nd quadrant.
∴ cot 1899° is negative.

iii. 986° = 2x 360° + 266°
∴ 986° and 266° are co-terminal angles.
Since 180° < 266° < 270°,
266° lies in the 3rd quadrant.
∴ 986° lies in the 3rd quadrant.
∴ sin 986° is negative.

Question 3.
State the quadrant in which 6 lies if
i. tan θ < 0 and sec θ > 0
ii. sin θ < 0 and cos θ < 0
iii. sin θ > 0 and tan θ < 0
Solution:
i. tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0
sec θ is positive in 1st and 4th quadrants.
∴ θ lies in the 4th quadrant.

ii. sin θ < 0
sin θ is negative in 3rd and 4th quadrants, cos θ < 0
cos θ is negative in 2nd and 3rd quadrants.
.’. θ lies in the 3rd quadrant.

iii. sin θ > 0
sin θ is positive in 1st and 2nd quadrants, tan θ < 0
tan θ is negative in 2nd and 4th quadrants.
∴ θ lies in the 2nd quadrant.

Question 4.
Which is greater?
sin (1856°) or sin (2006°)
Solution:
1856° = 5 x 360° + 56°
∴ 1856° and 56° are co-terminal angles.
Since 0° < 56° < 90°, 56° lies in the 1st quadrant.
∴ 1856° lies in the 1st quadrant,
∴ sin 1856° >0 …(i)
2006° = 5 x 360° + 206°
∴ 2006° and 206° are co-terminal angles.
Since 180° < 206° < 270°,
206° lies in the 3rd quadrant.
∴ 2006° lies in the 3rd quadrant,
∴ sin 2006° <0 …(ii)
From (i) and (ii),
sin 1856° is greater.

Question 5.
Which of the following is positive?
sin(-310°) or sin(310°)
Solution:
Since 270° <310° <360°,
310° lies in the 4th quadrant.
∴ sin (310°) < 0
-310° = -360°+ 50°
∴ 50° and – 310° are co-terminal angles.
Since 0° < 50° < 90°, 50° lies in the 1st quadrant.
∴ – 310° lies in the 1st quadrant.
∴ sin (- 310°) > 0
∴ sin (- 310°) is positive.

Question 6.
Show that 1 – 2sin θ cos θ ≥ 0 for all θ ∈ R.
Solution:
1 – 2 sin θ cos θ
= sin² θ + cos² θ – 2sin θ cos θ
= (sin θ – cos θ)² ≥ 0 for all θ ∈ R

Question 7.
Show that tan² θ + cot² θ ≥ 2 for all θ ∈ R.
Solution:

Question 8.
If sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\), then find the values of cos θ, tan θ in terms of x and y.
Solution:
Given, sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)
we know that
cos²θ = 1 – sin² θ

[Note: Answer given in the textbook of cos θ = \(\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = \(. However, as per our calculation the answer of cos θ = ± [latex]\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = ± \(\frac{x^{2}-y^{2}}{2 x y}\). ]

Question 9.
If sec θ = \(\sqrt{2}\) and \(\frac{3 \pi}{2}\) < θ < 2π, then evaluate \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\)
Solution:
Given sec θ = \(\sqrt{2}\)
We know that,
tan² θ = sec² θ – 1
= (\(\sqrt{2}\)) – 1
= 2 – 1 = 1
∴ tan θ = ±1
Since \(\frac{3 \pi}{2}\) < θ < 2π
θ lies in the 4th quadrant.
∴ tan θ < 0
∴ tan θ = -1

Question 10.
Prove the following:

i. sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1
Solution:
L.H.S. = sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A (cos² B + sin² B) + cos² A (sin² B + cos² B)
= sin²A(1) + cos²A(1)
= 1 = R.H.S.

ii. \(\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-{cosec}^{3} \theta}=\sin ^{2} \theta \cos ^{2} \theta\)
Solution:

iii. L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)\)
Solution:
L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}\)
= (tanθ + secθ)² + (tanθ – secθ)²
= tan² θ + 2 tan θ sec θ + sec² θ
+ tan² θ – 2 tan θ sec θ +.sec² θ
= 2(tan² θ + sec² θ)

iv. 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ
Solution:
LHS.
= 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ =  = 2 sec² θ – (sec² θ)² – 2cosec² θ + (cosec² θ)²
= 2(1+ tan² θ) – (1+ tan² θ)² – 2(1+ cot² θ)
+ (1+ cot² θ)²
= 2 + 2tan² θ – (1 + 2tan² θ + tan⁴ θ)
– 2 – 2cot² θ + 1 + 2cot² θ + cot⁴ θ
= 2 + 2.tan² θ – 1 – 2 tan² θ – tan⁴ θ – 2
– 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ
= cot⁴ θ – tan⁴ θ = R.H.S.

v. sin⁴ θ + cos⁴ θ = sin⁴ θ + cos⁴ θ
Solution:
L.H.S. = sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2sin² θ cos² θ
… [ v a² + b² = (a + b)² – 2ab]
= 1 – 2sin² θ cos² θ
= R.H.S.

vi. 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 = 0
L.H.S =
2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1=0
= sin⁶ θ + cos⁶ θ
= (sin² θ)³ + (cos² θ)³ = (sin² θ + cos² θ)³
– 3 sin² θ cos² θ (sin2 0 + cos2 0)
…[••• a³ + b³ = (a + b)³ – 3ab(a + b)]
= (1)³ – 3 sin² θ cos² θ(1)
= 1-3 sin² θ cos² θ sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2 sin² θ cos² θ
…[Y a² + b² = (a + b)² – 2ab]
= 1-2 sin² θ cos² θ
L.H.S.= 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2(1-3 sin² θ cos² θ) -3(1 – 2 sin² θ cos² θ) + 1
= 2-6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1 = c
= R.H.S.

vii. cos⁴ θ – sin⁴ θ + 1 = 2cos²θ
L.H.S. = cos⁴ θ – sin⁴ θ + 1
= (cos² θ)² – (sin² θ)² + 1 = (cos²θ + sin²θ) c(os² θ – sin²θ) +1
= (1) (cos²θ – sin²θ) + 1 = cos² θ + (1 – sin²θ)
= cos² θ + cos²θ = 2cos²θ = R.H.S.

viii. sin⁴θ + 2sin²θ cos²θ = 1 – cos⁴θ
L.H.S. = sin⁴θ + 2sin²θ cos²θ = sin²θ(sin²θ + 2cos²θ)
= (sin²θ) (sin²θ + cos²θ + cos²θ) = (1 – cos²θ) (1 + cos²θ)
= 1 – cos⁴θ = R.H.S.

ix. \(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta-\cos \theta}=2\)
Solution:

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ)
= 2 (sin² θ + cos² θ)
= 2(1)
= 2 = R.H.S.

x. tan² θ – sin² θ = sin⁴ θ sec² θ
Solution:
L.H.S. = tan² θ – sin² θ
= \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) – sin²θ
= sin² θ (\(\frac{1}{\cos ^{2} \theta}-1 \))
= \(\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\)
= (sin² θ) (sin² θ)sec² θ
= sin⁴ θ sec² θ
= R.H.S

xi. (sinθ + cosecθ)² + (cos θ + see θ)² = tan² θ + cot² θ + 7
Solution:
L.H.S. = (sinθ + cosecθ)² + (cos θ + see θ)²
= sin ² θ + cosec² θ + 2sinθ cosec θ
+ cos² θ + sec² θ + 2sec0 cos0
= (sin² θ + cos² θ) + cosec² θ + 2 + sec² θ + 2
= 1 + (1 + cot² θ) + 2 + (1 + tan² θ) + 2 = tan² θ + cot² θ + 7
= R.H.S.

xii. sin⁸θ – cos⁸θ = (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
Solution:
L.H.S. = sin⁸θ – cos⁸θ
= (sin⁴θ)² – (cos⁴θ)²
= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)
= [(sin² θ)² – (cos² θ)² ]
. [(sin² θ)² + (cos² θ)² ]
= (sin² θ + cos² θ) (sin² θ – cos² θ). [(sin² θ + cos² θ)² – 2sin² θ.cos² θ] …[Y a² + b² = (a + b)² – 2ab]
= (1) (sin² θ – cos² θ) (1² – 2sin² θ cos² θ)
= (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
= R.H.S.

xiii. sin⁶A + cos⁶A = 1 – 3 sin²A + 3sin⁴A
Soluiton:
L.H.S. = sin⁶A + cos⁶A
= (sin² A)³ + (cos² A)³
= (sin² A + cos² A)³
– 3sin²A cos²A(sin² A + cos² A)
…[ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3sin²A cos²A (1)
= 1 – 3sin²A cos²A
= 1 – 3 sin²A (1 – sin²A)
= 1 – 3 sin²A + 3sin⁴A
= R.H.S.

xiv. (1 + tanA tanB)² + (tanA – tanB)² = sec ²A sec²B
Solution:
L.H.S. = (1 + tanA tanB)² + (tanA – tanB)²
= 1 + 2tanA tanB + tan²A tan² + tan² A- 2tanA tanB + tan²B
= 1 + tan²A + tan² B + tan²A tan²B
= 1(1+ tan²A) + tan² B(1 + tan²A)
= (1 + tan²A) (1 + tan²B)
= sec²A sec²B = R.H.S.

xv. \(\frac{1+\cot \theta+{cosec} \theta}{1-\cot \theta+{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-1}{\cot \theta-{cosec} \theta+1}\)
Solution:
We know that cosec²θ – cot² θ = 1
∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

xvi. \(\frac{\tan \theta+\sec \theta-1}{\tan \theta+\sec \theta+1}=\frac{\tan \theta}{\sec \theta+1}\)
Solution:
We know that
tan²θ = sec² θ – 1
∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

xvii. \(\frac{{cosec} \theta+\cot \theta-1}{{cosec} \theta+\cot \theta+1}=\frac{1-\sin \theta}{\cos \theta}\)
Solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

xviii. \(\frac{{cosec} \theta+\cot \theta+1}{\cot \theta+{cosec} \theta-1}=\frac{\cot \theta}{{cosec} \theta-1}\)
solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 1
• Exercise 1.2 Class 11 Maths 1
• Miscellaneous Exercise 1 Class 11 Maths 1
• Exercise 2.1 Class 11 Maths 1
• Exercise 2.2 Class 11 Maths 1
• Miscellaneous Exercise 2 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

11th Maths Part 1 Trigonometry – I Exercise 2.2 Questions And Answers Maharashtra Board

Question 1.
If 2sin A = 1 = \(\sqrt{2}\) cos B and \(\frac{\pi}{2}\) < A < π, \(\frac{3 \pi}{2}\)
Solution:
Given, 2sin A = 1
∴ sin A = 1/2
we know that,
cos² A = 1 – sin² A = 1 – \(\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}\)
∴ cos A = \(\pm \frac{\sqrt{3}}{2}\)
Since \(\frac{\pi}{2}\) < A < π
A lies in the 2nd quadrant.

We know that,
Sin² B = 1 – cos² B = 1 – \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)\(\frac{1}{2}=\frac{1}{2}\)
∴ sin B = \(\pm \frac{1}{\sqrt{2}}\)
Since \(\frac{3 \pi}{2}\) < B < 2π
B lies in the 4th quadrant,

Question 2.
If \(\) and A, B are angles in the second quadran, then prove that 4cosA + 3 cos B = -5
Solution:
Given, \(\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}\)
∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)
We know that,
cos² A = 1 – sin² = 1 – \(\left(\frac{3}{5}\right)^{2}\) = 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ Cos A = ± \([{4}{5}\)
Since A lies in the second quadrant,
cos A < 0
∴ Cos A = –\(\frac{4}{5}\)
Sin B = 4/5
We know that,
cos²B = 1 – sin²B = 1 – \(\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}\)
∴ Cos B = ±\(\frac{4}{5}\)
Since B lies in the second quadrant, cos B < 0

Question 3.
If tan θ = \(\frac{1}{2}\), evaluate \(\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}\)
Solution:

Question 4.
Eliminate 0 from the following:
i. x = 3sec θ, y = 4tan θ
ii. x = 6cosec θ,y = 8cot θ
iii. x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
iv. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4tan θ,3y = 5 + 3sec θ
Solution:
i. x = 3sec θ, y = 4tan θ
∴ sec θ = \(\frac{x}{3}\) and tan θ= \(\frac{y}{4}\)
We know that,
sec²θ – tan²θ = 1

∴ 16x² – 9y² = 144

ii. x = 6cosec θ and y = 8cot θ
.’. cosec θ = \(\) and cot θ = \(\)
We know that,
cosec² θ – cot² θ =

16x² – 9y² = 576

iii. x = 4cos θ – 5 sin θ … (i)
y = 4sin θ + 5cos θ .. .(ii)
Squaring (i) and (ii) and adding, we get
x² + y² = (4cos θ – 5sin θ)² + (4sin θ + 5cos θ)²
= 16cos²θ – 40 sinθ cosθ + 25 sin²θ + 16 sin² θ + 40sin θ cos θ + 25 cos² θ
= 16(sin² θ + cos² θ) + 25(sin² θ + cos² θ)
= 16(1) + 25(1)
= 41

iv. x = 5 + 6cosec θ andy = 3 + 8cot θ
∴ x – 5 = 6cosec θ and y – 3 = 8cot θ
∴ cosec θ = \(\frac{x-5}{6}\) and cot θ = \(\frac{y-3}{8}\)
We know that,
cosec² θ – cot² θ = 1
∴ \(\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}\) = 1

v. 2x = 3 – 4tan θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ and 3y – 5 = 3sec θ
∴ tan θ = \(\frac{3-2 x}{4}\) and sec θ = \(\frac{3 y-5}{3}\)θ
We know that, sec² θ – tan² θ = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}\) = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}\) = 1

Question 5.
If 2sin² θ + 3sin θ = 0, find the permissible values of cosθ.
Solution:
2sin² θ + 3sin θ = 0
∴ sin θ (2sin θ + 3) = 0
∴ sin θ = 0 or sin θ = \(\frac{-3}{2}\)
Since – 1 ≤ sin θ ≤ 1,
sin θ = 0
\(\sqrt{1-\cos ^{2} \theta}\) = 0 …[ ∵ sin² θ = 1- cos² θ]
∴ 1 – cos² θ = 0
∴ cos² θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Question 6.
If 2cos² θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.
Solution:
2cos²θ – 11 cos θ + 5 = 0
∴ 2cos² θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2cos θ – 1 = 0
∴ cos θ = 5 or cos θ = 1/2
Since, -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

Question 7.
Find the acute angle θ such 2cos² θ = 3sin θ.
Solution:
2cos²0 = 3sin θ
∴ 2(1 – sin² θ) = 3sin θ
∴ 2 – 2sin² θ = 3sin θ
∴ 2sin² θ + 3sin 9-2 = θ
∴ 2sin² θ + 4sin θ – sin θ – 2 = θ
∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ
∴ (sin θ + 2) (2sin θ – 1) = 0
∴ sin θ + 2 = 0 or 2sin θ – 1 = 0
∴ sin θ = -2 or sin θ = 1/2
Since, -1 ≤ sin θ ≤ 1
∴ Sin θ = 1/2
∴ θ = 30° …[ ∵ sin 30 = 1/2]

Question 8.
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5tan² θ + 3 = 9sec θ
∴ 5(sec² θ – 1) + 3 = 9sec θ
∴ 5sec² θ – 5 + 3 = 9sec θ
∴ 5sec² θ – 9sec θ – 2 = 0
∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0
∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0
∴ (sec θ – 2) (5sec θ + 1) = 0
∴ sec θ – 2 = 0 or 5sec θ + 1 = 0
∴ sec θ = 2 or sec θ = -1/5
Since sec θ ≥ 1 or sec θ ≤ -1,
sec θ = 2
∴ θ = 60° … [ ∵ sec 60° = 2]

Question 9.
Find sin θ such that 3cos θ + 4sin θ = 4.
Solution:
3cos θ + 4sin θ = 4
∴ 3cos θ = 4(1 – sin θ)
Squaring both the sides, we get .
9cos²θ = 16(1 – sin θ)²
∴ 9(1 – sin² θ) = 16(1 + sin² θ – 2sin θ)
∴ 9 – 9sin² θ = 16 + 16sin² θ – 32sin θ
∴ 25sin² θ – 32sin θ + 7 = 0
∴ 25sin² θ – 25sin θ – 7sin θ + 7 = 0
25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0
∴ (sin θ – 1) (25sin θ – 7) = 0
∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0
∴ sin θ = 1 or sin θ = \(\frac{7}{25}\)
Since, -1 ≤ sin θ ≤ 1
∴ sin θ = 1 or \(\frac{7}{25}\)
[Note: Answer given in the textbook is 1. However, as per our calculation it is 1 or \(\frac{7}{25}\).]

Question 10.
If cosec θ + cot θ = 5, then evaluate sec θ.
Solution:
cosec θ + cot θ = 5
∴ \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5\)
∴ \(\frac{1+\cos \theta}{\sin \theta}=5\)
∴ 1 + cos θ = 5.sin θ
Squaring both the sides, we get
1 + 2 cos θ + cos² θ = 25 sin² θ
∴ cos² θ + 2 cos θ + 1 = 25 (1 – cos² θ)
∴ cos² θ + 2 cos θ + 1 = 25 – 25 cos² θ
∴ 26 cos² θ + 2 cos θ – 24 = 0
∴ 26 cos² θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ or 26 cos θ – 24 = 0
∴ cos θ = -1 or cos θ = \(\frac{24}{26}=\frac{12}{13}\)
When cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cos θ ≠ -1
∴ cos θ = \(\frac{12}{13}\)
∴ sec θ = \(\frac{1}{\cos \theta}=\frac{13}{12}\)
[Note: Answer given in the textbook is -1 or \(\frac{13}{12}\).
However, as per our calculation it is only \(\frac{13}{12}\).]

Question 11.
If cot θ = \(\frac{3}{4}\) and π < θ < \(\frac{3 \pi}{2}\), then find the value of 4 cosec θ + 5 cos θ.
Solution:
We know that,
cosec²θ = 1 + cot² θ = \(\left(\frac{3}{4}\right)^{2}\) = 1 + \(\frac{9}{16}\)
∴ cosec² θ = \(\frac{25}{16}\)
∴ cosec θ = \(\pm \frac{5}{4}\)
Since π < θ < \(\frac{3 \pi}{2}\)
θ lies in the third quadrant.
∴ cosec θ < 0
∴ cosec θ = –\(\frac{5}{4}\)
cot θ = \(\frac{3}{4}\)
tan θ = \(\frac{1}{\cot \theta}=\frac{4}{3}\)
We know that,
sec² θ = 1 + tan² θ = 1 + \(\left(\frac{4}{3}\right)^{2}\)
= 1 + \(\frac{16}{9}=\frac{25}{9}\)
∴ sec θ = ±\(\frac{5}{3}\)
Since θ lies in the third quadrant,
sec θ < 0
∴ sec θ = –\(\frac{5}{3}\)
cos θ = \(\frac{1}{\sec \theta}=\frac{-3}{5}\)
∴ 4cosec θ + 5cos θ
= \(4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)\)
= -5 – 3 = -8
[Note: The question has been modified.]

Question 12.
Find the Cartesian co-ordinates of points whose polar co-ordinates are:
i. (3, 90°) ii. (1, 180°)
Solution:
i. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 3cos 90° and y = 3sin 90°
∴ x = 3(0) = 0 and y = 3(1) = 3
∴ the required cartesian co-ordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 1(cos 180°) and y = 1(sin 180°)
∴ x = -1 and y = 0
∴ the required cartesian co-ordinates are (-1, 0).

Question 13.
Find the polar co-ordinates of points whose Cartesian co-ordinates are:
1. (5, 5) ii. (1, \(\sqrt{3}\))
ii. (-1, -1) iv. (-\(\sqrt{3}\), 1)
Solution:
i. (x, y) = (5, 5)
∴ r = \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{25+25}\)
\(=\sqrt{50}=5 \sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{5}{5}\) = 1
Since the given point lies in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar co-ordinates are (\(5 \sqrt{2}\), 45°).

ii. (x, y) = ( 1, \(\sqrt{3}\))
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}\)
Since the given point lies in the 1st quadrant,
θ = 60° …[∵ tan 60° = \(\sqrt{3}\)]
∴ the required polar co-ordinates are (2, 60°).

iii. (x, y) = (-1, -1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{-1}{-1}=1\)
∴ tan θ = tan \(\frac{\pi}{4}\)
Since the given point lies in the 3rd quadrant,
tan θ = tan \(\left(\pi+\frac{\pi}{4}\right)\) …[∵ tan (n + x) = tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{4}\right)\)
∴ θ = \(\frac{5 \pi}{4}\) = 225°
∴ the required polar co-ordinates are (\(\sqrt{2}\), 225°).

iv. (x, y) = (-\(\sqrt{3}\) , 1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{1}{-\sqrt{3}}\) = -tan \(\frac{\pi}{6}\)
Since the given point lies in the 2nd quadrant,
tan θ = tan \(\left(\pi-\frac{\pi}{6}\right)\) …[∵ tan (π – x) = – tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{6}\right)\)
∴ θ = \(\frac{5 \pi}{6}\) = 150°
∴ the required polar co-ordinates are (2, 150°)

Question 14.
Find the values of:
i. sin\(\frac{19 \pi^{e}}{3}\)
ii. cos 1140°
iii. cot \(\frac{25 \pi^{e}}{3}\)
Solution:
i. We know that sine function is periodic with period 2π.
sin \(\frac{19 \pi}{3}\) = sin \(\left(6 \pi+\frac{\pi}{3}\right)\) = sin \(\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

ii. We know that cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = \(\frac {1}{2}\)

iii. We know that cotangent function is periodic with period π.
cot \(\frac{25 \pi}{3}\) = cot \(\left(8 \pi+\frac{\pi}{3}\right)\) = cot \(\frac{\pi}{3}\) = \(\frac{1}{\sqrt{3}}\)
dhana work.txt
Displaying dhana work.txt.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 1
• Exercise 1.2 Class 11 Maths 1
• Miscellaneous Exercise 1 Class 11 Maths 1
• Exercise 2.1 Class 11 Maths 1
• Exercise 2.2 Class 11 Maths 1
• Miscellaneous Exercise 2 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.1 Questions and Answers.

11th Maths Part 1 Trigonometry – I Exercise 2.1 Questions And Answers Maharashtra Board

Question 1.
Find the trigonometric functions of 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, – 30°, – 45°, – 60°, – 90°, – 120°, – 225°, – 240°, – 270°, – 315°
Solution:
Angle of measure 0°:

Let m∠XOA = 0° = 0^c
Its terminal arm (ray OA) intersects the standard
unit circle in P(1, 0).
Hence,x = 1 and y = 0
sin 0° = y = 0,
cos 0° = x = 1,
tan 0° = \(\frac{y}{x}=\frac{0}{1}\) = 0
cot 0° = \(\frac{x}{y}=\frac{1}{0}\) which is not defined
sec 0° = \(\frac{1}{x}=\frac{1}{1}\) = 1
cot 0° = \(\frac{1}{y}=\frac{1}{0}\) which is not defined,

Angle of measure 30°:
Let m∠XOA = 30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y)
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{1}{\sqrt{2}}\) and
y = PM = \(\frac{1}{\sqrt{2}}\)
∴ P = (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\))

Angle of measure 60°:
Let m∠XOA = 60°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Angle of measure 150°:
Let m∠XOA = 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 180°:
Let m∠XOA = 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(-1, 0).
∴ x = – 1 and y = 0
sin 180° =y = 0
cos 180° = x = -1

tan 180° = \(\frac{y}{x}\)
= \(\frac{0}{-1}\) = 0
Cosec 180° = \(\frac{1}{y}\)
= \(\frac{1}{0}\)
which is not defined.
sec 180°= \(\frac{1}{x}=\frac{1}{-1}\) = -1
cot 180° = \(\frac{x}{y}=\frac{-1}{0}\) , which is not defined.

Angle of measure 210°:
Let m∠XOA = 210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 3rd quadrant, x < 0,y < 0
∴ x = -OM = \(\frac{-\sqrt{3}}{2}\) and y = -PM = \(\frac{-1}{2}\)
∴ P ≡( \(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\) )

Angle of measure 300°:
Let m∠XOA = 300°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0,y > 0
x = OM = \(\frac{1}{2}\) = and y = -PM = \(\frac{-\sqrt{3}}{2}\)
sin 300° = y = \(\frac{-\sqrt{3}}{2}\)
cos 300° = x = \(\frac{1}{2}\)
tan 300° = \(\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\)

Angle of measure 330°:
Let m∠XOA = 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 4th quadrant, x > 0, y < 0

Angle of measure 30°
Let m∠XOA = -30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60 — 90° triangle.
op = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

[Note : Answer given in the textbook of sin (45°) = – 1/2. However, as per our calculation it is \(-\frac{1}{\sqrt{2}}\) ]

Angle of measure (-60°):
Let m∠XOA = -60°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 4’ quadrant,
x > 0, y < 0
x = OM =\(\frac{1}{2}\) and y = -PM = \(-\frac{\sqrt{3}}{2}\)

Angle of measure (-90°):
Let m∠XOA = -90°
It terminal arm (ray OA) intersects the standard unit circle at P(0, -1)
∴ x = 0 and y = -1
sin (-90°) = y = -1
cos (-90°) = s = 0

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (- 225°):
Let m∠XOA = – 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 2400):
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30°  – 60° –  900 triangle.

Since point P lies in the 2nd quadrant, x<0, y>0

Angle of measure (- 270°):
Let m∠XOA = – 270°
Its terminal arm (ray OA)
intersects the standard unit,
circle at P(0, 1).
∴ x = 0 and y = 1
sin (- 270°) = y = 1
cos (- 270°) = x = 0
tan(-270°)= \(\frac{y}{x}=\frac{1}{0}\)
which is not defined.

Angle of measure ( 315°):
Let m∠XOA 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Question 2.
State the signs of:
i. tan 380°
ii. cot 230°
iii 468°
Solution:
1. 380° = 360° + 20°
∴ 380° and 20° are co-terminal angles.
Since 0° < 20° <90°0,
20° lies in the l quadrant.
∴ 380° lies in the 1st quadrant,
∴ tan 380° is positive.

ii. Since, 180° <230° <270°
∴ 230° lies in the 3rd quadrant.
∴ cot 230° is positive.

iii. 468° = 360°+108°
∴ 468° and 108° are co-terminal angles.
Since 90° < 108° < 180°,
108° lies in the 2nd quadrant.
∴ 468° lies in the 2nd quadrant.
∴ sec 468° is negative.

Question 3.
State the signs of cos 4^c and cos 4°. Which of these two functions is greater?
Solution:
Since 0° < 4° < 90°, 4° lies in the first quadrant. ∴ cos4° >0 …(i)
Since 1^c = 57° nearly,
180° < 4^c < 270°
∴ 4^c lies in the third quadrant.
∴ cos 4^c < 0 ………(ii)
From (i) and (ii),
cos 4° is greater.

Question 4.
State the quadrant in which 6 lies if
i. sin θ < 0 and tan θ > 0
ii. cos θ < 0 and tan θ > 0
Solution:
i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

Question 5.
Evaluate each of the following:
i. sin 30° + cos 45° + tan 180°
ii. cosec 45° + cot 45° + tan 0°
iii. sin 30° x cos 45° x lies tan 360°
Solution:
i. We know that,
sin30° = 1/2, cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 180° = 0
sin30° + cos 45° +tan 180°
= \(\frac{1}{2}+\frac{1}{\sqrt{2}}+0=\frac{\sqrt{2}+1}{2}\)

ii. We know that,
cosec 45° = \(\sqrt{2}\) , cot 45° = 1, tan 0° = 0
cosec 45° + cot 45° + tan 0°
= \(\sqrt{2}\) + 1 + 0 = \(\sqrt{2}\) + 1

iii. We know that,
sin 30° = \(\frac{1}{2}\), cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 360° = 0
sin 30° x cos 45° x tan 360°
= \(\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\) = 0

Question 6.
Find all trigonometric functions of angle in standard position whose terminal arm passes through point (3, – 4).
Solution:
Let θ be the measure of the angle in standard position whose terminal arm passes through P(3, -4).
∴ x = 3 and y = -4
r = OP

Question 7.
If cos θ = \(\frac{12}{13}, 0<\theta<\frac{\pi}{2}\) find the value of \(\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta}, \frac{1}{\tan ^{2} \theta}\)
Solution:
cos θ = \(\frac{12}{13}\)
We know that,
sin² θ = 1 – cos²θ

∴ sin θ = ± \(\frac{5}{13}\)
Since 0 < θ < \(\frac{\pi}{2}\) , θ lies in the 1st quadrant, ∴ sin θ > 0

Question 8.
Using tables evaluate the following:
i. 4 cot 45° – sec2 60° + sin 30°
ii.\(\cos ^{2} 0+\cos ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \frac{\pi}{2}\)
Solution:
i. We know that,
cot 45° = 1, sec 60° = 2, sin 30° = 1/2
4 cot 45° – sec² 60° + sin 30°
= 4(1) – (2)² + \(\frac{1}{2}\)
= 4 – 4 + \(\frac{1}{2}=\frac{1}{2}\)

ii. We know that,

Question 9.
Find the other trigonometric functions if
i. cot θ = \(-\frac{3}{5}\), and 180 < θ < 270
ii. Sec A = \(-\frac{25}{7}\) and A lies in the second quadrant.
iii cot x = \(\frac{3}{4}\), x lies in the third quadrant.
iv. tan x = \(\frac{-5}{12}\) x lies in the fourth quadrant.
Solution:
i. cot θ = \(-\frac{3}{5}\)
we know that,
sin²θ = 1 – cos²θ
= 1 – \(\left(-\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ sin θ = ± \(\frac{4}{5}\)
Since 180° < 0 < 270°,
θ lies in the 3rd quadrant.
∴ sin θ < 0

Since A lies in the 2nd quadrant,
tan A < 0

iii. Given, cot x = \(\frac{3}{4}\)
We know that,
cosec² x = 1 + cot² x
= 1 + \(\left(\frac{3}{4}\right)^{2}=1+\frac{9}{16}=\frac{25}{16}\)
∴ cosec x = ± \(\frac{5}{4}\)
Since x lies in the 3rd quadrant, cosec x < 0

iv. Given, tan x = \(-\frac{5}{12}\)
sec² x = 1 + tan²
= 1 + \(\left(-\frac{5}{12}\right)^{2}\)
= 1 + \(\frac{25}{144}=\frac{169}{144}\)
∴ sec x = ± \(\frac{13}{12}\)
Since x lies in the 4th quadrant,
sec x > 0

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 1
• Exercise 1.2 Class 11 Maths 1
• Miscellaneous Exercise 1 Class 11 Maths 1
• Exercise 2.1 Class 11 Maths 1
• Exercise 2.2 Class 11 Maths 1
• Miscellaneous Exercise 2 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Miscellaneous Exercise 1 Questions and Answers.

11th Maths Part 1 Angle and its Measurement Miscellaneous Exercise 1 Questions And Answers Maharashtra Board

I. Select the correct option from the given alternatives.

Question 1.
\(\left(\frac{22 \pi}{15}\right)^{c}x\) is equal to
(A) 246°
(B) 264°
(C) 224°
(D) 426°
Answer:
(B) 264°

Question 2.
156° is equal to

Answer:
(B)

Question 3.
A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 metres when it traces the angle of 12° at the centre, then the length of the rope is
(A) 70 m
(B) 55 m
(C) 40 m
(D) 35 m
Answer:
(A) 70 m

Question 4.
A pendulum 14 cm long oscillates through an angle of 12°, then the angle of the path described by its extremities is


Answer:
(D)

Question 5.
Angle between hands of a clock when it shows the time 9 :45 is
(A) (7.5)°
(B) (12.5)°
(C) (17.5)°
(D) (22.5)°
Answer:
(D) (22.5)°

Question 6.
20 metres of wire is available for fencing off a flower-bed in the form of a circular sector of radius 5 metres, then .the maximum area (in sq. m.) of the flower-bed is
(A) 15
(B) 20
(C) 25
(D) 30
Answer:
(C) 25
r + r + rθ = 20m
2r + rθ = 20

Question 7.
If the angles of a triangle are in the ratio 1:2:3, then the smallest angle in radian is
(A) \(\frac{\pi}{3}\)
(B) \(\frac{\pi}{6}\)
(C) \(\frac{\pi}{2}\)
(D) \(\frac{\pi}{9}\)
Answer:
(B) \(\frac{\pi}{6}\)

Question 8.
A semicircle is divided into two sectors whose angles are in the ratio 4:5. Find the ratio of their areas?
(A) 5:1
(B) 4:5
(C) 5:4
(D) 3:4
Answer:
(B) 4:5

Question 9.
Find the measure of the angle between hour- hand and the minute hand of a clock at twenty minutes past two.
(A) 50°
(B) 60°
(C) 54°
(D) 65°
Answer:
(A) 50°

Question 10.
The central angle of a sector of circle of area 9π sq.cm is 60°, the perimeter of the sector is
(A) π
(B) 3 + π
(C) 6 + π
(D) 6
Answer:
(C) 6 + π

II. Answer the following.

Question 1.
Find the number of sides of a regular polygon, if each of its interior angles is \(\frac{3 \pi^{c}}{4}\).
Solution:
Each interior angle of a regular polygon
= \(\frac{3 \pi}{4}=\left(\frac{3 \pi}{4} \times \frac{180}{\pi}\right)^{\circ}\) = 135°
Interior angle + Exterior angle = 180°
∴ Exterior angle = 180° – 135° = 45°
Let the number of sides of the regular polygon be n.
But in a regular polygon, exterior angle = \(\frac{360^{\circ}}{\text { no.of sides }}\)
∴ 45° = \(\frac{360^{\circ}}{\mathrm{n}}\)
∴ n = \(\frac{360^{\circ}}{45^{\circ}}\) = 8
∴ Number of sides of a regular polygon = 8.

Question 2.
Two circles each of radius 7 cm, intersect each other. The distance between their centres is 7√2 cm. Find the area common to both the circles.
Solution:
Let O and O₁ be the centres of two circles intersecting each other at A and B.
Then OA = OB = O₁A = O₁B = 7 cm
and OO₁ = 7√2 cm
OO₁² = 98 ………………(i)
Since OA² + O₁A² = 7²
= 98
= OO₁² …..[ from (i)]
m∠OAO₁ = 90°
□ OAO₁B is a square.
m∠AOB = m∠AO₁B = 90°

A(□ OAO₁B) = (side)² = (7)² = 49 sq.cm
∴ Required area = area of shaded portion = A(sector OAB) + A(sector O₁AB)) – A(□ OAO₁B)

Question 3.
∆PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle within the triangle.
Solution:
Let ‘O’ be the centre of the circle drawn on QR as a diameter.
Let the circle intersect seg PQ and seg PR at points M and N respectively.
Since l(OQ) = l(OM),
m∠OM Q = m∠OQM = 60°
m∠MOQ = 60°
Similarly, m∠NOR = 60°
Given, QR =18 cm.
r = 9 cm

θ = 60° = (60 x \(\frac{\pi}{180}\))^c
= \(\left(\frac{\pi}{3}\right)^{c}\)
∴ l(arc MN) = S = rθ = 9 x \(\frac{\pi}{3}\) = 3π cm.

Question 4.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm.
Solution:
Let S be the length of the arc and r be the radius of the circle.
θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
S = 37.4 cm
Since S = rθ,

Question 5.
A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees?
Solution:
S = 10 cm and r = 4 cm
Since S = rθ,
10 = 4 x θ

Question 6.
If two arcs of the same length in two circles subtend angles 65° and 110° at the centre. Find the ratio of their radii.
Solution:
Let r₁ and r₂ be the radii of the two circles and let their arcs of same length S subtend angles of 65° and 110° at their centres.
Angle subtended at the centre of the first circle,

Angle subtended at the centre of the second circle,

Question 7.
The area of a circle is 81TH sq.cm. Find the length of the arc subtending an angle of 300° at the centre and also the area of corresponding sector.
Solution:
Area of circle = πr²
But area is given to be 81 n sq.cm
∴ πr² = 81π
∴ r2 = 81
∴ r = 9 cm
θ = 300° = \(=\left(300 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{5 \pi}{3}\right)^{\mathrm{c}}\)
Since S = rθ
S = 9 x \(\frac{5 \pi}{3}\) = 15π cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 9 x 15π = \(\frac{135 \pi}{2}\) sq.cm

Question 8.
Show that minute-hand of a clock gains 5° 30′ on the hour-hand in one minute.
Solution:
Angle made by hour-hand in one minute
\(=\frac{360^{\circ}}{12 \times 60}=\left(\frac{1}{2}\right)^{\circ}\)
Angle made by minute-hand in one minute = \(\frac{360^{\circ}}{60}\) = 6°
∴ Gain by minute-hand on the hour-hand in one minute
= \(6^{\circ}-\left(\frac{1}{2}\right)^{\circ}=\left(5 \frac{1}{2}\right)^{\circ}\) = 5°30′
[Note: The question has been modified.]

Question 9.
A train is running on a circular track of radius 1 km at the rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.
Solution:
r = 1km = 1000m
l(Arc covered by train in 30 seconds)
= 30 x \(\frac{36000}{60 \times 60}\)m
∴ S = 300 m
Since S = rθ,
300 = 1000 x θ

= (17.18)°
= 17° +(0.18)°
= 17° + (0.18 x 60)’ = 17° + (10.8)’
∴ θ = 17°11′(approx.)

Question 10.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution:

Let ‘O’ be the centre of the circle and AB be the chord of the circle.
Here, d = 40 cm
∴ r = \(\frac{40}{2}\) = 20 cm
Since OA = OB = AB,
∆OAB is an equilateral triangle.
The angle subtended at the centre by the minor
arc AOB is θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
= l(minor arc of chord AB) = rθ = 20 x \(\frac{\pi}{3}\)
= \(\frac{20 \pi}{3}\) cm

Question 11.
The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radians.
Solution:
Let the measures of the angles of the quadrilateral in degrees be a – 3d, a – d, a + d, a + 3d, where a > d > 0
∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 360°
… [Sum of the angles of a quadrilateral is 360°]
∴ 4a = 360°
∴ a = 90°
According to the given condition, the greatest angle is double the least,
∴ a + 3d = 2.(a – 3d)
∴ 90° + 3d = 2.(90° – 3d)
∴ 90° + 3d = 180° – 6d 9d = 90°
∴ d = 10°
∴ The measures of the angles in degrees are
a – 3d = 90° – 3(10°) = 90° – 30° = 60°,
a – d = 90° – 10° = 80°,
a + d = 90°+ 10°= 100°,
a + 3d = 90° + 3(10°) = 90° + 30° = 120°

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 1
• Exercise 1.2 Class 11 Maths 1
• Miscellaneous Exercise 1 Class 11 Maths 1
• Exercise 2.1 Class 11 Maths 1
• Exercise 2.2 Class 11 Maths 1
• Miscellaneous Exercise 2 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Ex 1.2 Questions and Answers.

11th Maths Part 1 Angle and its Measurement Exercise 1.2 Questions And Answers Maharashtra Board

Question 1.
Find the length of an arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm.
Solution:
Here, r = 15cm and
θ = 108° = \(\left(108 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{3 \pi}{5}\right)^{\mathrm{c}}\)
Since S = r.θ
S = 15 x \(\frac{3 \pi}{5}\)
= 9π cm.

Question 2.
The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length equal to length of radius.
Solution:
Here, r = 9cm
Let the arc AB cut off a chord equal to the radius of the circle.
Since OA = OB = AB,
ΔOAB is an equilateral triangle.
m∠AOB = 60°
θ = 60°
= \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
Since S = r.θ,
S = 9 x \(\frac{\pi}{3}\) = 3π cm.

Question 3.
Find the angle in degree subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of the circle is 25 cm.
Solution:
Here, r = 25 cm and S = 15 cm
Since S = r.θ,
15 = 25 x θ

∴ The required angle in degree is \(\left(\frac{108}{\pi}\right)^{0}\) or (34.40)°(approx.).

Question 4.
A pendulum of length 14 cm oscillates through an angle of 18°. Find the length of its path.
Solution:

Question 5.
Two arcs of the same length subtend angles of 60° and 75° at the centres of the two circles. What is the ratio of radii of two circles?
Solution:
Let r₁, and r₂ be the radii of the two circles and let their arcs of same length S subtend angles of 60° and 75° at their centres.
Angle subtended at the centre of the first circle,
θ₁ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
∴ S = r₁θ₁ = r₁(\(\left(\frac{\pi}{3}\right)\))
Angle subtended at the centre of the second circle,

Question 6.
The area of the circle is 2571 sq.cm. Find the length of its arc subtending an angle of 144° at the centre. Also find the area of the corresponding sector.
Solution:
Area of circle = πr²
But area is given to be 25 π sq.cm
∴ 25π = πr²
∴ r² = 25
∴ r = 5 cm
θ = 144° = \(=\left(144 \times \frac{\pi}{180}\right)^{c}=\left(\frac{4 \pi}{5}\right)^{\mathrm{c}}\)
Since s = rθ
S = 5(\(\frac{4 \pi}{5}\)) = 4π
Also, A(sector) = \(\frac{1}{2}\) x r x S = \(\frac{1}{2}\) x 5 x 4π
= 10π sq. cm

Question 7.
OAB is a sector of the circle having centre at O and radius 12 cm. If m∠AOB = 45°, find the difference between the area of sector OAB and ΔAOB.
Solution:
Here, r = 12 cm
\(\theta=45^{\circ}=\left(45 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{4}\right)^{c}\)

Draw AM ⊥ OB
In ΔOAM,

[Note: The question has been modified.]

Question 8.
OPQ is the sector of a circle having centre at O and radius 15 cm. If m∠POQ = 30°, find the area enclosed by arc PQ and chord PQ.
Solution:
Here, r = 15 cm
m∠POQ = 30°
\(\left(30 \times \frac{\pi}{180}\right)^{c}[/larex]

∴ θ = [latex]\left(\frac{\pi}{6}\right)^{c}\)
Draw QM ⊥ OP
In ΔOQM,
sin 30° = \(\frac{\text { QM }}{15}\)
QM= 15 x \(\frac{1}{2}=\frac{15}{2}\)
Shaded portion indicates the area enclosed by arc PQ and chord PQ.
∴ A(shaded portion)
= A(sector OPQ) – A(ΔOPQ)

Question 9.
The perimeter of a sector of the circle of area 25π sq.cm is 20 cm. Find the area of sector.
Solution:
Area of circle = πr²
But area is given to be 25π sq.cm.
∴ 25π = πr²
∴ r² = 25
∴ r = 5 cm
Perimeter of sector = 2r + S
But perimeter is given to be 20 cm.
∴ 20 = 2(5) + S
∴ 20 = 10 + S
∴ S = 10 cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 5 x 10
= 25sq.cm.

Question 10.
The perimeter of a sector of the circle of area 64 7i sq.cm is 56 cm. Find the area of the sector.
Solution:
Area of circle = πr²
But area is given to be 25π sq.cm.
∴ 64π = πr²
∴ r² = 64
∴ r = 8 cm
Perimeter of sector = 2r + S
But perimeter is given to be 20 cm.
∴ 56 = 2(5) + S
∴ 56 = 16 + S
∴ S = 40 cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 8 x 40
= 160sq.cm.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 1
• Exercise 1.2 Class 11 Maths 1
• Miscellaneous Exercise 1 Class 11 Maths 1
• Exercise 2.1 Class 11 Maths 1
• Exercise 2.2 Class 11 Maths 1
• Miscellaneous Exercise 2 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Ex 1.1 Questions and Answers.

11th Maths Part 1 Angle and its Measurement Exercise 1.1 Questions And Answers Maharashtra Board

Question 1.
(A) Determine which of the following pairs of angles are co-terminal.
i. 210°, 150°
ii. 360°, -30°
iii. -180°, 540°
iv. -405°, 675°
v. 860°, 580°
vi. 900°, -900°
Solution:
210°,- 150°
210°-(- 150°) = 210°+ 150°
= 360°
= 1 (360°),
which is a multiple of 360°.
∴ The given pair of angles is co-terminal.

ii. 360°, – 30°
360° – (- 30°) = 360° + 30°
= 390°,
which is not a multiple of 360°.
∴ The given pair of angles is not co-terminal.

iii. -180°, 540°
540° -(-180°) = 540°+ 180°
= 720°
= 2(360°),
which is a multiple of 360°.
.’. The given pair of angles is co-terminal.

iv. – 405°, 675°
675° – (- 405°) = 675° + 405°
= 1080°
= 3(360°),
which is a multiple of 360°.
.’. The given pair of angles is co-terminal.

v. 860°, 580°
860° – 580° = 280°
which is not a multiple of 360, °.
∴ The given pair of angles is not co-terminal.

vi. 900°, 900°
900° – (-900°) = 900° + 900°
= 1800°
= 5(360°)
which is a multiple of 360°
∴ The given pair of angles is co-terminal.

Question 1.
(B) Draw the angles of the following measures and determine their quadrants.
i. -140°
ii. 250°
iii. 420°
iv. 750°
v. 945°
vi. 1120°
vii. – 80°
viii. – 330°
ix. – 500°
x. – 820°
Solution:

From the figure, the given angle terminates in quadrant III.

ii.

From the figure, the given angle terminates in quadrant III.

iii.

From the figure, the given angle terminates in quadrant I.

iv.

From the figure, the given angle terminates in quadrant I.

v.

From the figure, the given angle terminates in quadrant III.

vi.

From the figure, the given angle terminates in quadrant I.

vii.

From the figure, the given angle terminates in quadrant IV.

viii.

From the figure, the given angle terminates in quadrant I.

ix.

From the figure, the given angle terminates in quadrant III.
[Note: Answer given in the textbook is ‘Angle lies in quadrant II’. However, we found that it lies in quadrant III.]

x.

From the figure, the given angle terminates in quadrant III.

Question 2.
Convert the following angles into radians,
i. 85°
ii. 250°
iii. -132°
iv. 65°30′
v. 75°30′
vi. 40°48′
Solution:
we know that = \(\theta^{\circ}=\left(\theta \times \frac{\pi}{180}\right)^{c}\)
i. 85° = \(\left(85 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{17 \pi}{36}\right)^{\mathrm{c}}\)
ii. 250° = \(\left(250 \times \frac{\pi}{180}\right)^{c}=\left(\frac{25 \pi}{18}\right)^{c}\)
iii. 132° = \(\left(-132 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(-\frac{11 \pi}{15}\right)^{\mathrm{c}}\)
[Note : Answer given in the textbook is \(\frac{11 \pi}{15}\) However, as per our calculation it is \(\left(\frac{-11 \pi}{15}\right)^{c}\) ]

iv. 65° 30′ = 65° + 30′
= 65° + \(\left(\frac{30}{60}\right)^{\circ}\) … [1′ = (1/60)°]
= 65° + (1/2)°

v. 75° 30′ = 75° + 30′

vi. 40°48′ = 40° + 48′

Question 3.
Convert the following angles in degrees.
i. \(\frac{7 \pi^{c}}{12}\)
ii. \(\frac{-5 \pi^{c}}{3}\)
iii. 5^c
iv. \(\frac{11 \pi^{c}}{18}\)
v. \(\left(\frac{-1}{4}\right)^{c}\)
Solution:

Question 4.
Express the following angles in degrees, minutes and seconds.
i. (183.7)°
ii. (245.33)°
iii. \(\left(\frac{1}{5}\right)^{c}\)
Solution:
We know that 1° = 60′ and 1′ = 60″
i. (183.7)° = 183° +(0.7)°
= 183° + (0.7 x 60)’
= 183°+ 42′
= 183° 42′

ii. (245.33)° = 245° + (0.33)°
= 245° + (0.33 x 60)’
= 245° + (19.8)’
= 245° + 19’+ (0.8)’
= 245° 19’+ (0.8 x 60)”
= 245° 19’+ 48″
= 245° 19′ 48″

iii. We know that θ^c = (θ x \(\frac{180}{\pi}\))°

= (11.46)°
= 11° +(0.46)°
= 11° + (0.46×60)’
= 11°+ (27.6)’
= 11°+ 27’+ (0.6)’
= 11° + 27′ + (0.6×60)”
= 11°27′ + 36″
= 11°27’36” (approx.)

Question 5.
In △ABC, if m∠A = \(\frac{7 \pi^{\mathrm{c}}}{36}\), m∠B = 120°, find m∠C in degree and radian.
Solution:
We know that θ ^c = (θ x \(\left(\theta \times \frac{180}{\pi}\right)^{\circ}\) ) °
In △ABC,
m∠A = \(\frac{7 \pi^{\mathrm{c}}}{36}=\left(\frac{7 \pi}{36} \times \frac{180}{\pi}\right)^{\circ}\) = 35°
m∠B = 120°
∴ m∠A + m∠B + m∠C = 180°
… [Sum of the angles of a triangle is 180°]
∴ 35° + 120° + m∠C = 180° m∠C = 180° – 35° – 120°
∴ m∠C = 25°

Question 6.
Two angles of a triangle are \(\frac{5 \pi}{9}^{\mathrm{c}}\) and \(\frac{5 \pi}{18}^{\mathrm{c}}\) Find the degree and radian measures of third angle.
Solution:
We know that θ^c = [θ x \( ]°
The measures of two angles of a triangle are [latex]\frac{5 \pi^{\mathrm{c}}}{9}, \frac{5 \pi^{\mathrm{c}}}{18},\)
i.e., \(\left(\frac{5 \pi}{9} \times \frac{180}{\pi}\right)^{\circ},\left(\frac{5 \pi}{18} \times \frac{180}{\pi}\right)^{0}\)
i.e., 100°, 50°
Let the measure of third angle of the triangle be x°.
∴ 100°+50°+x° = 180°
…[Sum of the angles of a triangle is 180°]
∴ x° = 180°- 100° – 50°
∴ x° = 30°

∴ The degree and radian measures of the third angle are 30° and \(\left(\frac{\pi}{6}\right)^{\mathrm{c}}\) respectively.

Question 7.
In a right angled triangle, the acute angles are in the ratio 4:5. Find the angles of the triangle in degrees and radians.
Solution:
Since the triangle is aright angled triangle, one of the angles is 90°.
In the right angled triangle, the acute angles are in the ratio 4:5.
Let the measures of the acute angles of the triangle in degrees be 4k and 5k, where k is a constant.
∴ 4k + 5k+ 90°= 180°
… [Sum of the angles of a triangle is 180°]
∴ 9k = 180° – 90°
∴ 9k = 90°
∴ k = 10°
∴ The measures of the angles in degrees are
4k = 4 x 10° = 40°,
5k = 5 x 10° = 50°
and 90°.
we known that θ° = ( θ x \(\frac{\pi}{180}\)) ^c
∴ The measure of the angles in radius are

Question 8.
The sum of two angles is 5π^c and their difference is 60°. Find their measures in degrees.
Solution:
Let the measures of the two angles in degrees be x and y.
Sum of two angles is 5π^c
x + y = 5π^c
x + y = (5π x \( \frac{180}{\pi}\) ) …[ θ^c = \(\left(\theta \times \frac{180}{\pi}\right)^{\circ}\) ]
∴ x + y = 900° ………..(i)
∴ Difference of two angles is 60°.
x – y = 60° ….(ii)
Adding (i) and (ii), we get
2x = 960°
∴ x = 480°
Substituting the value of x in (i), we get
480° + y = 900°
∴ y = 900° — 480° = 420°
∴ The measures of the two angles in degrees are 480° and 420°.

Question 9.
The measures of the angles of a triangle are in the ratio 3:7:8. Find their measures in degrees and radians.
Solution:
The measures of the angles of the triangle are in the ratio 3:7:8.
Let the measures of the angles of the triangle in degrees be 3k, 7k and 8k, where k is a constant.
∴ 3k + 7k + 8k = 180°
… [Sum of the angles of a triangle is 180°]
∴ 18k =180°
∴ k = 10°
∴ The measures of the angles in degrees are
3k = 3 x 10° = 30°,
7k = 7 x 10° = 70° and
8k = 8 x 10° = 80°.

Question 10.
The measures of the angles of a triangle are in A.P. and the greatest is 5 times the smallest (least). Find the angles in degrees and radians.
Solution:
Let the measures of the angles of the triangle in degrees be a – d, a, a + d, where a > d> 0.
∴ a – d + a + a + d = 180°
…[Sum of the angles of a triangle is 180°]
∴ 3a = 180°
∴ a = 60° …(i)
According to the given condition, greatest angle is 5 times the smallest angle.
∴ a + d = 5 (a – d)
∴ a + d = 5a – 5d
∴ 6d = 4a
∴ 3d = 2a
∴ 3d = 2(60°) …[From (i)]
∴ d = \(\frac{120^{\circ}}{3}\) = 40°
∴ The measures of the angles in degrees are
a – d = 60° – 40° = 20°
a = 60° and
a + d = 60° + 40° = 100°

Question 11.
In a cyclic quadrilateral two adjacent angles are 40 and \(\frac{\pi^{c}}{3}\). Find the angles of the quadralateral in degrees.
Solution:
Let ABCD be the cyclic quadrilateral such that
∠A = 40° and

∴  ∠A + ∠C = 180°
∴ 40° + ∠C = 180°
∴ ∠C= 180°- 40°= 140°
Also, ∠B + ∠D = 180°
… [Opposite angles of a cyclic quadrilateral are supplementary]
∴ 60° + ∠D =180°
∴ ∠D = 180°- 60° = 120°
∴ The angles of the quadrilateral in degrees are 40°, 60°, 140° and 120°.

Question 12.
One angle of a quadrilateral has measure \(\frac{2 \pi^{c}}{5}\) and the measures of other three angles are in the ratio 2:3:4. Find their measures in degrees and radians.
Solution:
We know that θ^c = \(\left(\theta \times \frac{180}{\pi}\right)^{\circ}\))
One angle of the quadrilateral has measure\(\frac{2 \pi^{c}}{5}=\left(\frac{2 \pi}{5} \times \frac{180}{\pi}\right)^{\circ}=72^{\circ}\)
Measures of other three angles are in the ratio 2:3:4.
Let the measures of the other three angles of the quadrilateral in degrees be 2k, 3k, 4k, where k is a constant.
∴ 72° + 2k + 3k + 4k = 360°
…[Sum of the angles of a quadrilateral is 360°]
∴ 9k = 288°
k = 32°
∴ The measures of the angles in degrees are
2k = 2 x 32° = 64°
3k = 3 x 32° = 96°
4k = 4 x 32°= 128°
We know that θ° = (θ x \(\frac{\pi}{180}\))^c
∴ The measures of the angles in radians are

Question 13.
Find the degree and radian measures of exterior and interior angles of a regular
i. pentagon
ii. hexagon
iii. septagon
iv. octagon
Solution:
i. Pentagon:
Number of sides = 5
Number of exterior angles = 5
Sum of exterior angles = 360°

Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° — 72° = 108°
= \(

ii. Hexagon:
Number of sides = 6
Number of exterior angles = 6
Sum of exterior angles = 360°

Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° – 60° = 120°
= (120 x [latex]\frac{\pi}{180}\))^c = ( \(\frac{2 \pi}{3}[latex])^c

iii. Septagon:
Number of sides = 7
Number of exterior angles = 7
Sum of exterior angles = 360°
∴ Each exterior angle = [latex]\frac{360^{\circ}}{\text { no. of sides }}=\frac{360^{\circ}}{7}\)
= (51.43)°
= \(\left(\frac{360}{7} \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{2 \pi}{7}\right)^{\mathrm{c}}\)
Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° – ( \(\frac{360}{7}\))°

iv. Octagon:
Number of sides = 8
Number of exterior angles = 8
Sum of exterior angles = 360°
∴ Each exterior angle = \(\frac{360^{\circ}}{\text { no. of sides }}=\frac{360^{\circ}}{8}\)
= 45°
= \(\left(45 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{4}\right)^{c}\)
Interior angle + Exterior angle = 180°
Each interior angle = 180° – 45° = 135°
= \(\left(135 \times \frac{\pi}{180}\right)^{c}=\left(\frac{3 \pi}{4}\right)^{c}\)

Question 14.
Find the angle between hour-hand and minute-hand in a clock at
i. ten past eleven
ii. twenty past seven
iii. thirty five past one
iv. quarter to six
v. 2:20
vi. 10:10
Solution:
i. At 11:10, the minute-hand is at mark 2 and hour-hand has crossed \(\left(\frac{1}{6}\right)^{\text {th }}\) of the angle between 11 and 12.

Angle between two consecutive marks = \(\frac{360^{\circ}}{12}\) = 30°
Angle traced by hour-hand in 10 minutes
= \(\frac{1}{6}\) (30°) = 5°
Angle between marks 11 and 2 = 3 x 30° = 90°
∴ Angle between two hands of the clock at ten past eleven = 90° – 5° = 85°

ii. At 7 : 20 the minute -hand is at mark 4 and hour -hand has crossed \(\left(\frac{1}{3}\right)^{ }\)rd of angle between 7 and 8.

Angle between two consecutive marks
= 360°/12 = 30°
Angle traced by hour-hand in 20 minutes
= \(\frac{1}{3}\)(30°)= 10°
Angle between marks 4 and 7 = 3 x 30° = 90°
Angle between two hands of the clock at twenty past seven = 90° – 10° = 100°

iii. At 1 : 35 the minute -hand is at mark 7 and hour -hand has crossed \(\left(\frac{7}{12}\right)^{ }\)th of angle between 1 and 2.

Angle between two consecutive marks
= 360°/12 = 30°
Angle traced by hour-hand in 35 minutes
= \(\frac{7}{12}\left(30^{\circ}\right)=\left(\frac{35}{2}\right)^{\circ}=\left(17 \frac{1}{2}\right)^{\circ}\frac{1}{3}\)
Angle between marks 1 and 7 = 6 x 30° = 180°
Angle between two hands of the clock at thirty five past one = 180° – \(\left(17 \frac{1}{2}\right)^{\circ}=\left(162 \frac{1}{2}\right)^{\circ}\)
= 162° + \(\frac{1}{2}\) = 162°30′

iv. At 5:45, the minute-hand is at mark 9 and hour- hand has crossed ( \(frac{3}{4}\) )th of the angle between 5 and 6.

Angle between two consecutive marks
= 360°/12 = 30°
Angle traced by hour-hand in 45 minutes
\(\frac{3}{4}\left(30^{\circ}\right)=(22.5)^{\circ}=\left(22 \frac{1}{2}\right)^{\circ}\)
Angle between marks 5 and 9
= 4 x 30° = 120°
∴ Angle between two hands of the clock at quarter to six = \(120^{\circ}-\left(22 \frac{1}{2}\right)^{0}\)

v. At 2 : 20, the minute-hand is at mark 4 hour hand has crossed \(\frac{1}{3}\)rd of the angle between 2 and 3.

Angle between two consecutive marks = 360°/12 = 30°
Angle traced by hour-hand in 20 minutes
= \(frac{1}{3}\)(30°)= 10°
Angle between marks 2 and 4 = 2 x 30° = 60°
∴ Angle between two hands of the clock at 2 :20 = 60° – 10° = 50°

vi. At 10:10, the minute-hand is at mark 2 and hour-hand has crossed\frac{1}{6}[/latex] th between 10 and 11.

Angle between two consecutive marks
360°/12 = 30°
Angle traced by hour-hand in 10 minutes
= \(\frac{1}{6}\) (30°) = 5°
Angle between marks 10 and 2= 4 x 30° = 120°
… Angle between two hands of the clock at 10:10
= 120° – 5°= 115°

Class 11 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 11 Maths 1
• Exercise 1.2 Class 11 Maths 1
• Miscellaneous Exercise 1 Class 11 Maths 1
• Exercise 2.1 Class 11 Maths 1
• Exercise 2.2 Class 11 Maths 1
• Miscellaneous Exercise 2 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Economics Solutions Chapter 10 Economic Planning in India Textbook Exercise Questions and Answers.

Std 11 Economics Chapter 10 Question Answer Economic Planning in India Maharashtra Board

Class 11 Economics Chapter 10 Economic Planning in India Question Answer Maharashtra Board

Economics Class 11 Chapter 10 Question Answer Maharashtra Board

1. Choose the correct option:

Questions 1.
Statements that are true about the Planning Commission:
(a) Planning Commission was established in 1950.
(b) The Prime Minister is the Ex-Officio Chairman of the Planning Commission.
(c) Economic planning is a time-bound program.
(d) Economic planning is based on predetermined objectives.
Options:
(1) a and b
(2) a, b, c, and d
(3) a and c
(4) None of these
Answer:
(1) a and b

Question 2.
Statements that are incorrect regarding India’s five-year plans:
(a) The main objective of the first five-year plan was the development of agriculture.
(b) Social welfare and poverty eradication were the prime objectives of the seventh five-year plan.
(c) By the second five-year plan, focus increased on faster, inclusive growth.
(d) Development of both agriculture and industry were the main objectives of the third five-year plan.
Options:
(1) a
(2) a, b and d
(3) c
(4) b and d
Answer:
(3) c

Question 3.

Group – ‘A’	Group – ‘B’
1. Economic planning	(a) Selection by Prime Minister
2. Twelfth Five Year Plan	(b) Think Tank group of Indian Government
3. NITI Aayog	(c) Fast and sustainable growth
4. NITI Aayog Vice-Chairperson	(d) Time-bound programme

Options:
(1) 1 – c, 2 – a, 3 – d, 4 – b
(2) 1 – d, 2 – b, 3 – a, 4 – c
(3) 1 – d, 2 – c, 3 – b, 4 – a
(4) 1 – b, 2 – d, 3 – c, 4 – a
Answer:
Correct pair: (3) 1 – d, 2 – c, 3 – b, 4 – a

Question 4.
Choose the correct statement:
Statement 1 – NITI Aayog takes note of the dynamic change in the Indian economy.
Statement 2 – Considering the economic, social, and technological differences in underdeveloped districts, the body plans to implement various programmes and bring about economic changes.
Options:
(a) statement 1 is correct
(b) statement 2 is correct
(c) statement 2 is the result of statement 1
(d) there is no relation between statement 1 and statement 2.
Answer:
(c) statement 2 is the result of statement 1

2. Give economic terms:

Question 1.
The conscious and deliberate choice of economic priorities by some public authority.
Answer:
National Agenda

Question 2.
A group of people called forth by the government to discuss various problems and also try to find solutions to them.
Answer:
Think Tank

3. Identify and explain the concepts from the given illustrations:

Question 1.
Sayali’s mother maintains a book of accounts for household purposes and plans the expenditure accordingly.
Answer:
Planning.
It is a time-bound programme. The objectives should fulfill by using available resources within the time limit.

Question 2.
Ramabai gets a subsidy on domestic LPG directly transferred to her bank a/c.
Answer:
Service Delivery.
It is the important target of the 12th five-year plan. The main aim behind this policy is to prevent corruption.

Question 3.
To solve classroom-related issues, the teacher forms a group of students. This group discusses the problems and finds solutions to them.
Answer:
Think-Tank

• It is important to function of NITI Aayog to solve the problems of our country.
• Think-Tank is a group of experts who are discussing and solve various problems of India.

4. Answer the following:

Question 1.
Explain the features of economic planning.
Answer:
Economic planning is a time-bound programme to achieve certain objectives by allocating available resources under the control of a central planning authority.

Prof. H. D. Dickinson defines economic planning as – “Economic planning is the making of major economic decisions such as what and how much is to be produced how, when and where it is to be produced, to whom it is to be allocated, by the conscious decision of the determinate authority, on the basis of a comprehensive survey of the economy as a whole.”

On the basis of the above definition, its main characteristics are:

• Central Planning Authority (CPA): There is a central planning authority that formulates the plans. In India, this authority is known as Planning Commission.
• Survey: There is a complete survey of the economy regarding the availability and use of natural and human resources.
• Objectives: It lays down certain objectives which are realistic and flexible.
• Priorities: Priorities are fixed according to the importance of each sector for its development.
• Mobilization of resources: Resources are mobilised through various sources like taxation, deficit financing, savings, etc.
• Plan period: Each plan is for a specific period, usually five years.
• Evaluation: From time to time, an assessment of the plan objectives is done to make changes if necessary.
• Continuous process: Economic planning is a continuous process which aims at the economic development of a country.
• Co-ordination: In India, economic planning is implemented by the Centre and State Governments together.
• Flexibility: There is flexibility in India’s economic planning so it’s possible to make changes as per the need.

Question 2.
Explain the targets of the 12th Five Year plan.
Answer:
Targets for infrastructure are:

• To connect all villages of a country with all-weather roads.
• To increase rural television and telephone density to 70%.
• To increase infrastructure investment to 9% of G.D.P.
• To upgrade national and state highways to a minimum two-lane standard.
• To achieve real GDP growth rate at 8%, agriculture growth rate at 4%, and manufacturing and industrial growth rate at 10 %.
• To reduce the headcount ratio of poverty by 10%.
• To create 50 million (5 crores) new work opportunities in the non-farm sector.
• To increase average years of schooling to 7 years.
• To eliminate gender and social gap in school enrollment.
• To reduce the total fertility rate to 2.1%.

Question 3.
Explain the structure of NITI Aayog.
Answer:
The structure of NITI Aayog includes Governing Council, Regional Councils, Special Invitees, and Organisational Framework. The organizational framework includes Chairperson, Vice-Chairperson, Ex-officio members, CEO, and Secretariat.

Question 4.
Explain the functions of NITI Aayog.
Answer:
Functions of NITI Aayog:

• To evolve a shared vision of national development, priority sector, and strategies with the active involvement of states in the light of national objectives.
• To act as ‘Best Friend at the Centre’.
• To formulate plans at the village level and aggregate higher levels of government.
• To provide feedback for constant innovative improvements.
• To provide advice and encourage partnership with national and international Think-Tank.
• To create a knowledge, innovation, and entrepreneurial support system.
• To offer a platform for the resolution of inter-sectoral and inter-departmental issues.
• To maintain a state-of-the-art resource center for research on good governance.
• To focus on technology up-gradation and capacity building.
• To foster Cooperative federalism, with the active involvement of states.

Question 5.
Distinguish between Planning Commission and NITI Aayog.
Answer:

NITI Aayog	Planning Commission
(i) It serves as an advisory Think Tank.	(i) It served as the extra-constitutional body.
(ii) It draws membership from wider expertise.	(ii) It had limited expertise.
(iii) It serves in the spirit of cooperative federalism as states are equal partners.	(iii) States participated as spectators in the annual plan meetings.
(iv) Secretaries to be known as CEO appointed by Prime Minister.	(iv) Secretaries were appointed through the usual process.
(v) It focuses upon the Bottom-up approach.	(v) It followed a Top-down approach.
(vi) It does not process mandate to impose policies.	(vi) It imposed policies on states and tied allocation of funds with projects it approved.
(vii) It does not have powers to allocate funds, which are vested in the finance minister.	(vii) It had powers to allocate funds to ministers and state government.
(viii) It was established on 1st January 2015.	(viii) It was established on 15th March 1950.

5. State with reasons whether you agree or disagree with the following statements:

Question 1.
State governments have a more significant role to play under NITI Aayog.
Answer:
Yes, I do agree with the statement.

• Under the planning commission, there was a one-way flow of policy i.e. from, Central Government to State Government.
• In NITI Aayog, the flow of policy is from Central Government to State Government and State Government to Central Government between ministers.
• In the policy-making of the Central Government, State Government will work closely with the center.
• Governing Council of NITI Aayog consists of all Chief Ministers of the States.
• Thus, State Government has a more significant role to play under NITI Aayog.

Question 2.
Functions of the Planning Commission have been transferred to NITI Aayog.
Answer:
Yes, I do agree with the statement.

• The planning commission enjoyed the powers to allocate funds to ministry and state government.
• NITI Aayog/Think Tank is an advisory body that performs the function of allocation of funds.
• On 31st January 2015, Planning Commission was replaced by NITI Aayog for making policies and to implement them.
• NITI Aayog aimed at expanding the role of the states, making the role of the state stronger in collaboration with the center.

Question 3.
The objective of the 12th five-year plan was to achieve faster, sustainable and inclusive growth.
Answer:
Yes, I do agree with the statement.

• 12th five-year plan aims at a GDP growth rate of 8%.
• It seeks to achieve 4% growth in the agriculture sector.
• If aimed to generate 50 million work opportunities in the non-farm sector and providing skill certification.
• Connecting all the villages of the country with all-weather roads.
• Increasing green cover by 1 million hectares every year.
• These objectives will help the country to achieve faster, sustainable and inclusive growth.

6. Read the following passage and answer the questions given below:

The Finance Minister of the Central Government presents the Union Budget before the Parliament during the month of February every year. The budget, also referred to as the annual financial statement reflects the estimated receipts and expenditure of the government for a particular financial year that begins on the 1st of April and ends on 31 sc March. Changes in the tax structure are suggested in the budget. Besides this, provisions are also made for allocating expenditure on defense, education, research, and development, etc. The date for presenting the budget has been shifted to the 1st of February every year. This enables the generation of funds well in advance prior to the commencement of the financial year.

Question 1.
Where is the Union Budget usually presented?
Answer:
Union Budget is presented in the Parliament.

Question 2.
What all aspects are considered while preparing the budget?
Answer:
Changes in the tax structure, provisions for allocating expenditure on defense, education, research, and development.

Question 3.
Why is the date for presenting the budget shifted to the 1st of February?
Answer:
The date for presenting the budget is shifted to the 1st of February because it enables the generation of funds well in advance prior to the commencement of the financial year.

Question 4.
Explain the term ‘budget’.
Answer:
A budget is the annual financial statement that shows estimated receipts and expenditures of the government for a year.

11th Economics Digest Chapter 10 Economic Planning in India Intext Questions and Answers

Collect information on: (Textbook Page No. 64)

1. Bombay Plan
2. People’s Plan
3. Gandhian Plan

Answer:

1. Bombay Plan: It is the name given to a World War II era. It is a set of proposals for the development of the post-independence economy of India.
2. Peoples Plan: It is to provide satisfaction to the immediate basic needs of India within a period of ten years.
3. Gandhian Plan: Espousing with the spirit of Gandhian economic thinking, Shriman Narayan Agarwal formulated this plan in 1944. This plan laid more emphasis on agriculture.

Find out: (Textbook PageNo. 65 )

Calculate D₆ and D₉ from the above table using ‘achievements’ as the numerical data.

Answer:

(i) Sixth Decile:
D₆ = size of 6 \(\left(\frac{n+1}{10}\right)^{t h}\) Observation
= size of 6 \(\left(\frac{11+1}{10}\right)^{t h}\) Observation
= size of 6 \(\left(\frac{12}{10}\right)^{t h}\) Observation
= size of 6 (1.2)th Observation
= size of (7.2)th Observation
size of 7.2 th 0bservation lies in of 7.7
∴ D₆ = 2

(ii) Ninth Decile:
D₉ = size of 9 \(\left(\frac{n+1}{10}\right)^{t h}\) Observation
= size of 9 \(\left(\frac{11+1}{10}\right)^{t h}\) Observation
= size of 9 \(\left(\frac{12}{10}\right)^{t h}\) Observation
= size of 9 (1.2)th Observation
= size of 10.8th Observation
size of 10.8 th 0bservation lies in of 13.7.
∴ D₉ = 4

Find out: (Textbook Page No. 66)

Information of various levels of National Family Health Survey (NFHS).
Answer:

• NFHS – 1: The first NFHS was conducted in 1992-93. The survey collected extensive information on population, health, and nutrition, with an emphasis on women and children.
• NFHS – 2: The second NFHS was conducted in 1998-99 in all states (26) of India. It was on the quality of health and family planning services, domestic violence, reproductive health, anemia, etc.
• NFHS – 3: The third NFHS was conducted in 2005-06, in 29 states of India. UNICEF, USAID, DFID, USA, providing funds and technical help for NFHS-3.
• NFHS – 4: The fourth NFHS was conducted in 2014-2015. USA was a major financial supporter for NFHS-4. It was conducted in 29 states and 6 union territories and focused on 640 districts in the country.

Do you know? (Textbook Page No. 66)

Think-tank: Think-tank is a group of experts who are gathered together by an organization, especially by a Government in order to consider various problems, try and work out ways to solve them.
Answer:

• ‘Think-Tank’ is a group of experts who come together, to form an organization.
• They study the various problems of an economy and try to bring solutions to solve those problems.
• It is counted under the premier policy of the Government of India.
• Its main objective is to find a shared vision of national development with the active participation of the states.
• It provides guidance to foster ‘cooperative federalism in the states.

Find out: (Textbook Page No. 67)

The present structure of NITI Aayog and list out the names of members in the respective columns.

Answer:

Chairperson	Vice-Chairperson	Members
Prime Minister	Rajiv Kumar	Ex-Officio Members, Special Invitees, Full-time Members

11th Std Economics Questions And Answers:

• Basic Concepts in Economics Class 11 Economics Questions And Answers
• Money Class 11 Economics Questions And Answers
• Partition Values Class 11 Economics Questions And Answers
• The Economy of Maharashtra Class 11 Economics Questions And Answers
• Rural Development in India Class 11 Economics Questions And Answers
• Population in India Class 11 Economics Questions And Answers
• Unemployment in India Class 11 Economics Questions And Answers
• Poverty in India Class 11 Economics Questions And Answers 
• Economic Policy of India Since 1991 Class 11 Economics Questions And Answers
• Economic Planning in India Class 11 Economics Questions And Answers

Balbharti Maharashtra State Board Class 11 Economics Solutions Chapter 9 Economic Policy of India Since 1991 Textbook Exercise Questions and Answers.

Std 11 Economics Chapter 9 Question Answer Economic Policy of India Since 1991 Maharashtra Board

Class 11 Economics Chapter 9 Economic Policy of India Since 1991 Question Answer Maharashtra Board

Economics Class 11 Chapter 9 Question Answer Maharashtra Board

1. Complete the following statements by choosing the correct alternative:

Question 1.
After Independence, India had adopted ____________
(a) Socialism
(b) Capitalism
(c) Mixed Economy
(d) Communism
Answer:
(c) Mixed economy

Question 2.
The new economic policy approved foreign technology in ____________
(a) Cottage industries
(b) Small scale industries
(c) Micro enterprises
(d) High priority industries
Answer:
(d) High priority industries

Question 3.
At present, the number of industries reserved for public sector has been reduced to ____________
(a) 3
(b) 5
(c) 7
(d) 2
Answer:
(d) 2

2. Assertion and Reasoning questions:

Question 1.
Assertion (A): Delicensing of industries was an important step taken under liberalization.
Reasoning (R): Unwanted controls and restrictions led to economic stagnation prior to 1991.
(a) (A) is TRUE but (R) is FALSE
(b) (A) is FALSE but (R) is TRUE
(c) (A) and (R) both are TRUE and (R) is the correct explanation of (A)
(d) (A) and (R) both are TRUE but (R) is not the correct explanation of (A)
Answer:
(c) (A) and (R) both are TRUE and (R) is the correct explanation of (A)

Question 2.
Assertion (A): In 1990-91, India faced an acute shortage of foreign exchange reserves.
Reasoning (R): Import quotas and tariffs led to an increase in imports.
(a) (A) is TRUE but (R) is FALSE
(b) (A) is FALSE but (R) is TRUE
(c) (A) and (R) both are TRUE and (R) is the correct explanation of (A)
(d) (A) and (R) both are TRUE but (R) is not the correct explanation of (A)
Answer:
(a) (A) is TRUE but (R) is FALSE

Question 3.
Assertion (A): Post liberalization, the sale of domestic goods has increased.
Reasoning (R): The demand for imported goods had increased due to liberal policy.
(a) (A) is TRUE but (R) is FALSE
(b) (A) is FALSE but (R) is TRUE
(c) (A) and (R) both are TRUE and (R) is the correct explanation of (A)
(d) (A) and (R) both are TRUE but (R) is not the correct explanation of (A)
Answer:
(b) (A) is FALSE but (R) is TRUE

Question 4.
Assertion (A): Due to Globalisation, a country cannot achieve self-sufficiency in food production.
Reasoning (R): Globalisation has created a revolution in the IT sector.
(a) (A) is TRUE but (R) is FALSE
(b) (A) is FALSE but (R) is TRUE
(c) (A) and (R) both are TRUE and (R) is the correct explanation of (A)
(d) (A) and (R) both are TRUE but (R) is not the correct explanation of (A)
Answer:
(d) (A) and (R) both are TRUE but (R) is not the correct explanation of (A)

3. Find the odd word out:

Question 1.
New Economic Policy – Liberalization, Privatization, Demonetization, Globalisation
Answer:
Demonetization

Question 2.
Industries requiring compulsory licensing – defense equipment, agro-based industries, cigarettes, industrial explosives
Answer:
agro-based industries

Question 3.
Navratna status companies – SPCL, IOC, ONGC, HPCL
Answer:
SPCL

Question 4.
Liberalization dealt with the following – MRTP, FERA, SEBI, NTPC
Answer:
NTPC

4. Identify and explain the concepts from the given illustrations:

Question 1.
Vehicles manufactured by various automobile companies are now available in India.
Answer:
Globalization.
Globalization means the interaction of the domestic economy with the rest of the world with regard to foreign investment, trade, production, and financial matters.

Question 2.
Government equity in some public sector enterprises is sold to the private sector.
Answer:
Disinvestment.
A disinvestment is an act of selling shares of sick public sector units to the private sector.
Eg. Disinvestment of Maruti, ITDC hotels, VSNL, etc.

Question 3.
Foreign investments are encouraged on a large scale in the industrial sector of India.
Answer:
Foreign Direct Investment (FDI).
FDI was approved under the Industrial Policy of 1991, to encourage investment in high-priority industries which require high investment and technology.

5. State with reasons whether you agree or disagree with the following statements:

Question 1.
Liberalization has permitted the use of foreign technology.
Answer:

• Yes, I do agree with the statement.
• Liberalization has encouraged foreign technology.
• Foreign technology is allowed in high-priority industries.
• Foreign technology helps to reduce the cost and make the industries competitive.

Question 2.
The government has given private enterprises free access to the public sector.
Answer:
Yes, I do agree with the statement.

• 17 industries were reserved for the public sector under the Industrial Policy of 1956.
• But in NEP – 1991, the number of public sector industries reduced from 17 to 2.
• Railway transport and atomic energy are reserved for the public sector.
• The involvement of the private sector in economic activities has increased after NEP.

Question 3.
Government has a monopoly in the insurance sector.
Answer:
No, I do not agree with the statement.

• The insurance sector was a monopoly of the Government till 1991.
• In 1991IRDA (Insurance Regulatory and Development Authority Act) was introduced.
• The IRDA has given licenses to many private companies to start insurance businesses in India.
• Due to the entry of private companies, the monopoly of government has come to an end.

Question 4.
The creation of the National Renewal Board (NRB) was done to remove poverty.
Answer:
Yes, I do agree with the statement.

• Under the public sector, some units were closed due to loss.
• The workers of these units had to face the problem of unemployment and poverty.
• To solve this problem, the government has been formed. National Renewal Board (NRB).
• NRB provides compensation to retrenched workers which help to reduce poverty in the country.

Question 5.
Indian Oil Corporation is one of the public sector units among ‘Navratnas’.
Answer:
Yes, I do agree with the statement.

• Navratnas are the Public Sector Units (PSUs).
• In 1997-98, Nine PSUs were selected for Navratna status.
• These PSUs were selected on the basis of their performance.
• These Navratnas were given full financial and managerial autonomy.

6. Answer in detail:

Question 1.
Explain the features of the New Economic Policy of 1991.
Answer:
The process of the new economic policy started in 1985 and got momentum in 1991.

Features of Economic Policy, 1991:

• Delicensing: The new industrial policy abolished all industrial licensing, except 18 specified industries related to security and strategic concerns and social reasons.
• Abolition of MRTP Act: No prior approval of the MRTP commission is now required for setting up industrial units by the large business houses.
• Encouragement to Small Scale Industries (SSI): The investment limit of the SSI has been increased up to 5 crores which will help to upgrade their machinery.
• Encouraging Foreign Investment: Many industrial units were open to foreign investment under the 1991 policy. The limit was raised to 51% and 100% in some industries and 100% in mining, pollution control equipment, electricity generation projects, ports, etc.
• Reducing the role of the Public Sector: The number of industries reserved for the public sector was reduced from 17 to 2, it includes railways and atomic energy.
• Trade Liberalisation: Relaxation is given to importers by abolishing import licensing controls. The permission for external credit and set up of Special Economic Zones (SEZ) to promote export. To promote agricultural export Agro Export Zones (AEZ) were introduced.
• Reforms in Insurance Sectors: The Insurance Regulatory and Development Authority Act (IRDA) has given licenses to many private companies to start insurance businesses which ended the monopoly of government e.g. Max Life, Bajaj, Allianz, Aegon, etc.
• Reforms in Financial Sector: The NEP has allowed private banks and foreign banks to do hanking business in the financial sector.

Question 2.
Explain the measures undertaken for Globalisation.
Answer:
Globalization means the interaction of the domestic economy with the rest of the world with regard to foreign investment, trade, production, and financial matters.

Measures were taken for Globalisation:

• Removal of quantitative restrictions: To make the Indian economy attractive to foreign investors, the government has reduced custom duties and tariffs imposed on imports and exports.
• Encouragement to foreign capital: To India, foreign investment has wider scope since 1991. Foreign capital is allowed in India without any restrictions.
• Convertibility of Rupee: It means Indian currency can be converted into the currency of other countries.
• Foreign collaboration: To take the benefit of advanced technology, Indian companies are allowed to enter into foreign collaboration e.g. Maruti-Suzuki, Hero-Honda, etc.
• Long-term trade policy: The trade policy was introduced for a longer duration to promote foreign trade.
• Encouragement to export: Many incentives have been given to industries through EXIM policy. SEZ and AEZ are created to encourage export.

7. Read the following passage carefully and answer the questions:

The Indian ice cream industry is one of the fastest-growing segments of the dairy and food processing sector. India has a low per capita consumption of ice cream of 400 ml whereas in the USA it is 22,000 ml and in China, it is 3000ml.

The per capita consumption of ice cream is low in India because it is a country filled with traditional sweets of more than 100 varieties. In developed countries, people have either pastries or ice-creams for dessert. In the era of globalization, the mindset of the people is fast changing. This is because multi-national companies have set up a number of ice-cream parlors, with a lot more varieties and flavours that attract the younger lot. Besides this, there are better delivery systems.

The ice-cream sector has great potential for growth in the country due to improvement in the cold chain infrastructure, increasing disposable income, and changing the lifestyle of the people. However, it is taxed higher with 18 percent GST while other dairy products in the same basket such as butter and cheese are taxed at 12 percent.

The ice-cream industry has generated revenue of more than $1.5 billion in 2016-17. With the employment of 15 lakh people directly or indirectly, it is also considered one of the largest employers of the dairy and food processing industry.

Question 1.
Identify the reason for the low per capita consumption of ice cream in India.
Answer:
In India, traditional sweets are available, which are having more than 100 varieties.

Question 2.
Explain the impact of globalisation on the Indian ice-cream industry.
Answer:
Due to globalisation, multinational companies have set up a number of ice-cream parlours with a lot of varieties and flavours. It helps to attract the younger generation of today.

Question 3.
Find out the factors that could lead to the growth of the ice-cream industry in India.
Answer:
In India Ice-cream industry has wider scope because there is an improvement in cold chain infrastructure, increase in disposable income, and changing lifestyle of the people.

Question 4.
Express your views about the implications of higher GST on the ice-cream industry in India.
Answer:
The ice-cream sector is indirectly dependent on the primary sector. If the demand for ice cream increased then the income of cattle owners will grow. 18% GST on ice cream is high because ice cream is made from milk which is good for health as compared to tobacco, pan masala. Same GST (18%) is imposed on tobacco and pan masala but it is injurious to health and finally, the burden of GST will transfer to customers.

11th Economics Digest Chapter 9 Economic Policy of India Since 1991 Intext Questions and Answers

Find out: (Textbook Page No. 58)

Names of five Private Banks and Foreign Banks.
Answer:

• Private Banks – ICICI Bank, Axis Bank, Kotak Mahindra Bank, Yes Bank, HDFC Bank.
• Foreign Banks – Standard Chartered Bank, DBS Bank, Doha Bank, Bank of America, Royal Bank of Scotland.

Find out: (Textbook Page No. 60)

Names of companies coming under Maharatna and Miniratna status.
Answer:
Maharatna

• Coal India Ltd.
• Gas Authority of India (GAIL)
• Indian Oil Corporation Ltd (IOCL)
• Mahanagar Telephone Nigam Ltd. (MTNL)

Miniratna

• Airports Authority of India
• Bharat Earth Movers Ltd.
• Bharat Dynamics Ltd.
• Mazagon Dock Ltd.
• State Trading Corporation of India

Stimulate your memory: (Textbook Page No. 61)

What is Corporate Social Responsibility (CSR)? How does it help society?
Answer:

• CSR means whatever a company does to give back to the community in which it has a presence.
• It is the company’s effort to improve society and the environment in some way.
• It helps society by providing education, healthcare, disaster relief measures, economic empowerment, planting trees, maintaining parks, etc.
• E.g. Mahindra & Mahindra constructed 4340 toilets in 104 districts of India, especially for girls in Government schools in 2013-14.

Activity-Based Questions

Observe the chart and answer the following question.

Question 1.
What is globalisation?
Answer:
Globalisation is a process of integrating the domestic economy with the rest of the world with regard to foreign investment, trade, production, and financial matters.

Question 2.
Explain the concept of disinvestment.
Answer:
Disinvestment is a process of selling shares of sick Public Sector Units (PSUs) to the private sector, so as to increase the production activities of that units and to achieve efficiency in the allocation of resources, improvement in management, etc.
E.g. Disinvestment of Maruti, ITDC hotels. IPCL. VSNL, etc.

Question 3.
Write the full form of FERA, SEZ, AEZ.
Answer:

• FERA – Foreign Exchange Regulation Act
• SEZ – Special Economic Zones
• AEZ – Agro Export Zones

Question 4.
Why NRB is created?
Answer:
National Renewal Board was created to look after the retrenched workers who become unemployed due to the closure of loss-making Public Sector Units (PSUs). Through this Board, the government took the responsibility of providing compensations to the retrenched workers and also to take care of those seeking voluntary retirement

Question 5.
By what was FERA replaced and why?
Answer:
FERA was replaced by FEMA to encourage international trade and to bring flexibility in the laws relating to foreign exchange.

11th Std Economics Questions And Answers:

• Basic Concepts in Economics Class 11 Economics Questions And Answers
• Money Class 11 Economics Questions And Answers
• Partition Values Class 11 Economics Questions And Answers
• The Economy of Maharashtra Class 11 Economics Questions And Answers
• Rural Development in India Class 11 Economics Questions And Answers
• Population in India Class 11 Economics Questions And Answers
• Unemployment in India Class 11 Economics Questions And Answers
• Poverty in India Class 11 Economics Questions And Answers 
• Economic Policy of India Since 1991 Class 11 Economics Questions And Answers
• Economic Planning in India Class 11 Economics Questions And Answers