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Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 5
Chapter Name	Trigonometric Ratios
Exercise	5.2
Category	RD Sharma Solutions

Table of Contents

RD Sharma Solutions for Class 10 Chapter 5 Trigonometric Ratios Ex 5.2 Download PDF

Chapter 5: Trigonometric Ratios Exercise – 5.2

Question: 1

sin 45° sin 30° + cos 45° cos 30°

Solution:

Sin 45°sin 30° + cos 45° cos 30° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 2

sin 60° cos 30° + cos 60° sin 30°

Solution:

sin 60° cos 30° + cos 60° sin 30° [1]

By trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 3

cos 60° cos 45° – sin 60° sin 45°

Solution:

cos 60° cos 45° – sin 60° sin 45° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 4

sin²30° + sin²45° + sin²60° + sin²90°

Solution:

sin²30° + sin²45° + sin²60° + sin²90° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 5

cos²30° + cos²45° + cos²60° + cos²90°

Solution:

cos²30° + cos²45° + cos²60° + cos²90° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 6

tan²30° + tan²45° + tan²60°

Solution:

tan²30° + tan²45° + tan²60° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 7

2 sin²30° − 3 cos²45° + tan²60°

Solution:

2sin²30° − 3cos²45° + tan²60° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 8

sin²30° cos²45°+ 4 tan²30°+ 1/2sin² 90° – 2 cos² 90°+ 1/24 cos²0°

Solution:

sin²30° cos²45°+ 4 tan²30°+ 1/2sin² 90° – 2 cos² 90°+ 1/24 cos²0° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 9

4(sin⁴60° + cos⁴30°) − 3(tan²60° − tan²45°) + 5cos²45°

Solution:

4(sin⁴60° + cos⁴30°) − 3(tan²60° − tan²45°) + 5cos²45° [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 10

(cosec²45° sec²30°) (sin²30° + 4 cot²45° − sec²60°)

Solution:

(cosec²45° sec²30°)(sin²30° + 4 cot²45° − sec²60°) [1]

We know that by trigonometric ratios we have,

Substituting the values in equation 1, we get

Question: 11

cosec³30°cos 60° tan³45° sin²90° sec²45°cot 30°

Solution:

= cosec³30°cos 60° tan³45° sin²90° sec²45° cot 30°

Question: 12

cot²30° − 2cos²60° − 3/4 sec²45° – 4 sec²30°

Solution:

= cot²30° − 2cos²60° − 3/4 sec²45° – 4 sec²30°

Question: 13

(cos 0° + sin 45° + sin 30°) (sin 90^∘ − cos 45° + cos 60°)

Solution:

Given, (cos 0° + sin 45° + sin 30°) (sin 90^∘ − cos 45° + cos 60°)

Question: 14

Solution:

Given,

Question: 15

Solution:

Given,

Question: 16

4(sin⁴30° + cos²60°) − 3(cos²45° − sin²90°) − sin²60°

Solution:

Given,

4(sin⁴30° + cos²60°) − 3(cos²45° − sin²90°) − sin²60°

Question: 17

Solution:

Given,

= 3 + 2 + 4

= 9

Question: 18

Solution:

Given,

Question: 19

Solution:

Question: 20

2 sin 3x = √3

Solution:

Given,

2 sin 3x = √3

⟹ sin 3x = √3/2

⟹ sin 3x = sin 60°

⟹ 3x = 60°

⟹ x = 20°

Question: 21

Solution:

x/2 = 30°

x = 60^°

Question: 22

√3 sin x = cos x

Solution:

√3 tan x = 1

tan x = 1/√3

∴ tan x = tan45°

x = 45^°

Question: 23

tan x = sin 45^° cos 45^° + sin 30^°

Solution:

tan x = 1

tan x = 45^°

x = 45^°

Question: 24

√3 tan 2x = cos 60°+ sin 45°cos 45°

Solution:

2x = 30^°

x = 15^°

Question: 25

cos 2x = cos 60° cos 30° + sin 60° sin 30°

Solution:

2x = 30°

x = 15°

Question: 26

If θ = 30°, verify

(iv) cos 3θ = 4cos³θ − 3cos θ cos 3θ = 4 cos³θ – 3 cos θ

Solution:

Substitute θ = 30° θ = 30° in equation (i)

Therefore, LHS = RHS

Substitute θ = 30° θ = 30°

Therefore, LHS = RHS.

Substitute θ = 30°θ = 30°

Therefore, LHS = RHS

(iv) cos 3θ = 4cos³θ − 3cos θ cos 3θ = 4 cos³θ – 3 cos θ

Solution:

LHS = cos 3θ

Substitute θ = 30°

= cos 3(30^°) = cos 90^°

= 0

RHS = 4 cos³θ – 3 cos θ

= 4 cos³30° − 3 cos 30°

= 0

Therefore, LHS = RHS.

Question: 27

If A = B = 60^°. Verify

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

Solution:

cos (A – B) = cos A cos B + sin A sin B … (i)

Substitute A and B in (i)

⟹ cos (60^° – 60^°) = cos 60^° cos 60^° + sin 60^° sin 60^°

⟹ 1 = 1

Therefore, LHS = RHS

(ii) Substitute A and B in (i)

⟹ sin (60^° – 60^°) = sin 60^° cos 60^° – cos 60^° sin 60^°

⟹ sin0^° = 0

⟹ 0 = 0

Therefore, LHS = RHS

A = 60^°, B = 60^° we get,

tan 0° = 0

0 = 0

Therefore, LHS = RHS

Question: 28

If A = 30^°, B = 60^° verify:

(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:

(i) A = 30^°, B = 60^° we get

Sin (30^° + 60^°) = Sin 30^° Cos 60^° + Cos 30^° Sin 60^°

Sin (90^°) = 1 ⟹ 1 = 1

Therefore, LHS = RHS

(ii) Cos (A + B) = Cos A Cos B – Sin A Sin B

A = 30^°, B = 60^° we get

Cos (30^° + 60^°) = Cos 30^° Cos 60^° – Sin 30^° Sin 60^°

0 = 0

Therefore, LHS = RHS

Question: 29

If sin(A + B) = 1 and cos(A – B) = 1, 0° < A + B ≤ 90°, A ≥ B find A and B.

Solution:

Given,

sin(A + B) = 1 this can be written as sin (A + B) = sin (90°) sin (90°)

cos(A – B) = 1 this can be written as cos (A – B) = cos (0°) cos (0°)

⟹ A + B = 90°

A – B = 0°

2A = 90°

A = 90°/2

A = 45°

Substitute A value in A – B = 0°

45°– B = 0°

B = 45°

Hence, the value of A = 45°and B = 45°

Question: 30

If tan (A – B) = 1/√3 and tan (A + B) = √3, 0°<A + B ≤ 90°, A > B find A and B

Solution:

Given,

A – B = 30° … 1

A + B = 60° … 2

Solve equations 1 and 2

A + B = 30°

A – B = 60°

2A = 90°

A = 90°/2

A = 45°

Substitute the value of A in equation 1

45°+ B = 30°

B = 30° – 45°

B = 15°

The value of A = 45°and B = 15°

Question: 31

If sin (A – B) = 1/2 and cos (A + B) = 1/2, 0°<A + B ≤ 90°, A < B. find A and B.

Solution:

Given,

A + B = 60° … 2

Solve equations 1 and 2

A + B = 60°

A – B = 30°

2A = 90°

A = 90°/2

A = 45°

Substitute the value of A in equation 2

45°+ B = 60°

B = 60° – 45°

B = 15°

The value of A = 45°and B = 15°

Question: 32

In a Δ ABC right angled triangle at B, ∠A = ∠C. Find the values of:

1. sin A cos C + cos A sin C

2. sin A sin B + cos A cos B

Solution:

1. since, it is given as ∠A = ∠C

the value of A and C is 45°, the value of angle B is 90°

because the sum of angles of triangle is 180°

⟹ sin 45° cos 45° + cos 45° sin 45°

⟹ 1

The value of sin A cos C + cos A sin C is 1

2. since, it is given as ∠A = ∠C

the value of A and C is 45°, the value of angle B is 90°

because the sum of angles of triangle is 180°

⟹ sin 45°sin 90° + cos 45°sin 90°

The value of sin A sin B + cos A cos B is 1/√2

Question: 33

Find the acute angle A and B, if sin (A + 2B) = √3/2 and cos(A + 4B) = 0, A > B.

Solution:

Given,

A + 2B = 60° … 1

Cos (A + 4B) = 0

A + 4B = sin⁻¹(90) sin⁻¹(90)

A + 4B = 90° … 2

Solve equations 1 and 2

2B = 30°

B = 30°/2

B = 15°

Substitute B value in equation 2

A + 4B = 90°

A + 4(15°) = 90°

A + 60° = 90°

A = 90° – 60°

A = 30°

The value of A = 30° and B = 15°

Question: 34

In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠ P and ∠ R.

Solution:

In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm

By Pythagoras theorem,

PR²= PQ² + QR²

⟹ 6² = 3² + QR²

⟹ QR² = 36 – 9

⟹ QR = √27

⟹ QR = 3√3

sin R = 3/6 = 1/2 = sin30°

∠R = 30°

As we know, Sum of angles in a triangle = 180

∠P + ∠Q + ∠R = 180°

⟹ ∠P + 90° + 30° = 180°

⟹ ∠P = 180° – 120°

⟹ ∠P = 60°

Therefore, ∠R = 30°

And, ∠P = 60°

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:

Given,

sin (A – B) = sin A cos B – cos A sin B

And, cos (A – B) = cos A cos B + sin A sin B

We need to find, sin 15 and cos 15.

Let A = 45 and B = 30

sin 15 = sin (45- 30) = sin 45 cos 30 – cos 45 sin 30

cos 15 = cos (45- 30) = cos 45 cos 30 – sin 45 sin 30

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:

Sin 60°= x/15

cos 60°= x/15

x = 7.5 units

Question: 37

In ΔABC is a right triangle such that ∠C = 90°, ∠A = 45°and BC = 7 units. Find the remaining angles and sides.

Solution:

Here, ∠C = 90° and ∠A = 45°

We know that,

∠A + ∠B + ∠C = 180°

⟹ 45°+ 90° + ∠C = 180°

⟹ 135° + ∠C = 180°

⟹ ∠C = 180° – 135°

⟹ ∠C = 45°

The value of the remaining angle C is 45°

Now, we need to find the sides x and y here,

y =7√2 units

x = 7 units

the value of x = 7 units and y = √2 units

Question: 38

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

Solution:

Let AC = x cm and CB = y cm

⇒ x = 40 cm = AC

Similarly BD = 40 cm

Now,

Question: 39

If A & B are acute angles such that tan A = 1/2 tan B = 1/3 and tan (A + B) =find A+B.

Solution:

(A + B) = Tan^-1 (1)

(A + B) = 45^°

Question: 40

Prove that: (√3 -1) (3 – cot 30°) = tan³60° – 2 sin 60°

Solution:

R.H.S ⟹ tan³60° – 2 sin 60°

L.H.S = R.H.S

Hence prove

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