RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.2

RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.2

RD Sharma Class 10 Solutions Trigonometric Ratios Exercise 5.2

Evaluate each of the following (1-19) : 1.
Question 1.
sin45° sin30° + cos45° cos30°.
Solution:
sin 45° sin 30° + cos 45° cos 30°

Question 2.
sin60° cos30° + cos60° sin30°.
Solution:

Question 3.
cos60° cos45° – sin60° sin45°.
Solution:

Question 4.
sin ² 30° + sin ² 45° + sin ² 60° + sin ² 290°.
Solution:

Question 5.
cos ² 30° + cos ² 45° + cos ² 60° + cos ² 90°.
Solution:

Question 6.
tan ² 30° + tan ² 60° + tan ² 45°.
Solution:

Question 7.
2sin ² 30° – 3cos ² 45° + tan ² 60°.
Solution:

Question 8.
sin ² 30°cos ² 4S° + 4tan ² 30° + \(\frac { 1 }{ 2 }\) sin ² 90° -2cos ² 90° + \(\frac { 1 }{ 24 }\) cos20°.
Solution:

Question 9.
4 (sin ⁴ 60° + cos ⁴ 30°) – 3 (tan ² 60° – tan ² 45°) + 5cos ² 45°
Solution:

Question 10.
(cosecc ² 45° sec ² 30°) (sin ² 30° + 4cot ² 45° – sec ² 60°).
Solution:

Question 11.
cosec ³ 30° cos 60° tan ³ 45° sin2 90° sec ² 45° cot 30°.
Solution:

Question 12.
cot ² 30° – 2cocs ² 60° – \(\frac { 3 }{ 4 }\)sec2 45° – 4sec ² 30°.
Solution:

Question 13.
(cos0° + sin45° + sin30°) (sin90° – cos45° + cos60°)
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
4 (sin ⁴ 30° + cos ² 60°) – 3 (cos ² 45° – sin ² 90°) – sin ² 60°
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Find the value of x in each of the following : (20-25)

Question 20.
2sin 3x = √3
Solution:

Question 21.
2sin \(\frac { x }{ 2 }\) = 1
Solution:

Question 22.
√3 sin x=cos x
Solution:

Question 23.
tan x = sin 45° cos 45° + sin 30°
Solution:

Question 24.
√3 tan 2x = cos 60° +sin 45° cos 45°
Solution:

Question 25.
cos 2x = cos 60° cos 30° + sin 60° sin 30°
Solution:

Question 26.
If θ = 30°, verify that :

Solution:

Question 27.
If A = B = 60°, verify that:
(i) cos (A – B) = cos A cos B + sin A sin B
(ii) sin (A – B) = sin A cos B – cos A sin B tan A – tan B
(iii) tan (A – B) = \(\frac { tanA-tanB }{ 1+tanA-tanB }\)
Solution:

Question 28.
If A = 30° and B = 60°, verify that :
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B – sin A sin B
Solution:

Question 29.
If sin (A + B) = 1 and cos (A,-B) = 1,0° < A + B < 90°, A > B find A and B.
Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
In a ∆ABC right angle at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Solution:

Question 33.
Find acute angles A and B, if sin (A + 2B)=\(\frac { \sqrt { 3 } }{ 2 }\) and cos (A + 4B) = 0, A > B.
Solution:

Question 34.
In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Solution:

Question 35.
If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Solution:

Question 36.
In a right triangle ABC, right angled at ∠C if ∠B = 60° and AB – 15 units. Find the remaining angles and sides.
Solution:

Question 37.
If ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Solution:

Question 38.
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Solution:

Question 39.
If A and B are acute angles such that

Solution:

Question 40.
Prove that (√3 + 1) (3 – cot 30°) = tan ³ 60° – 2sin 60°. [NCERT Exemplar]
Solution:

RD Sharma Class 10 Solutions Chapter 5 Trigonometric Ratios Ex 5.2

RD Sharma Class 10 Solutions

• Chapter 5 Trigonometric ratios Ex 5.1
• Chapter 5 Trigonometric ratios Ex 5.3