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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 10 Sine and Cosine Formulae and Their Applications – Exercise 10.1 | Set 1

### Question 1: If in △ABC, ∠A=45°, ∠B=60°, and ∠C=75°, find the ratio of its sides.

Solution:

According to the sine rule

Hence, we get

Using the formula,

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

sin (45°+30°) = sin(45°) cos(30°) + cos(45°) sin(30°)

sin 75° =

sin 75° =

Multiplying the denominator by 2√2, we get

Hence, we get

a : b : c = 2 : √6 : (√3+1)

### Question 2: If in △ABC, ∠C=105°, ∠B=45°, and a=2, then find b.

Solution:

In the given condition, ∠C=105° and ∠B=45°

And as we know that,

A+B+C = π (Sum of all angles in triangle is supplementary)

A = π-(B+C)

A = 180°-(45°+105°)

A = 30°

Now, according the sine rule

After rationalizing the denominator, we get

### Question 3: In △ABC, if a=18, b=24, and c=30 and ∠C=90°, find sin A, sin B, and sin C.

Solution:

In the given condition, a=18, b=24 and c=30 and ∠C=90°

Now, according the sine rule

Also,

### Question 4: In △ABC, prove the following:

Solution:

According to the sine rule

Considering the LHS of the equation, we have

By using trigonometric formula,

sin A – sin B = 2 sincos

sin A + sin B = 2 sincos

As, LHS = RHS

Hence, proved !!

### (a-b)= c

Solution:

According to the sine rule

Considering LHS, we have

By using trigonometric formula,

sin A – sin B = 2 sincos

= λ

A+B+C = π (Sum of all angles in triangle is supplementary)

= 2λ sincoscos

= 2λ sincoscos

= λ sin(2cossin)

By using trigonometric formula,

2 sin a cos a = sin 2a

= λ sin(sin)

= λ sinsin(A+B)

= λ sinsin(π-C) (A+B+C = π)

= λ sinsin(C)

= (λ sin C) sin

= c sin

As, LHS = RHS

Hence Proved!

### Question 6: In △ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

By using trigonometric identities,

sin A – sin B = 2 sincos

sin 2a = 2 sin a cos a

……………………….(1)

Now considering RHS, we have

Cross multiplying we get,

By using trigonometric identities,

sin a cos b + cos a sin b = sin (a+b)

sin a cos b – cos a sin b = sin (a-b)

……………………….(2)

As, LHS = RHS

Hence Proved!

### Question 7: In △ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

By using trigonometric identities,

sin A + sin B = 2 sincos

sin 2a = 2 sin a cos a

………………………….(1)

Now considering RHS, we have

Cross multiplying we get,

By using trigonometric identities,

cos a cos b + sin a sin b = cos (a-b)

cos a cos b – sin a sin b = cos (a+b)

……………………….(2)

As, LHS = RHS

Hence, Proved!

### Question 8: In △ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

By using trigonometric identities,

sin A + sin B = 2 sincos

sin 2a = 2 sin a cos a

As, LHS = RHS

Hence, Proved!

### Question 9: In △ABC, prove the following:

Solution:

According to the sine rule

Considering RHS, we have

By using trigonometric identities,

sin A – sin B = 2 sincos

By using trigonometric identities,

2 sin a cos a = sin 2a

As, LHS = RHS

Hence, Proved!

### Question 10: In △ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

By using trigonometric identities,

sin2 a – sin2 b = sin(a+b) sin (a-b)

As, LHS = RHS

Hence, Proved!

### b sin B – c sin C = a sin (B-C)

Solution:

According to the sine rule

And, cosine rule,

Considering RHS, we have

RHS = a sin (B-C)

By using trigonometric identities,

sin(a-b) = sin a cos b – cos a sin b

= a (sin B cos C – cos B sin C)

= a ((bλ)(cλ))

= λ

= 2λ

= λb2-λc2

= b(λb) – c(λc)

= b(sin B) – c(cos C)

As, LHS = RHS

Hence, Proved!

### a2 sin (B-C) = (b2-c2) sin A

Solution:

According to the sine rule

Considering RHS, we have

RHS = (b2-c2) sin A

= λ2 sin A(sin2 B – sin2 C)

By using trigonometric identities,

sin2 a – sin2 b = sin(a+b) sin (a-b)

= λ2 sin A(sin(B+C) sin (B-C))

= λ2 sin A(sin(\pi-A) sin (B-C))

= λ2 sin A(sin(A) sin (B-C))

= λ2 sin2 A sin (B-C)

= (λ sin A)2 sin (B-C)

= a2 sin (B-C)

As, LHS = RHS

Hence, Proved!

### Question 13: In △ABC, prove the following:

Solution:

Considering LHS, we have

LHS =

Rationalizing the denominator, we get

According to the sine rule

As, LHS = RHS

Hence, Proved!

### a (sin B – sin C) + b(sin C – sin A) + c(sin A – sin B) = 0

Solution:

According to the sine rule

Considering LHS, we have

LHS = a (sin B – sin C) + b(sin C – sin A) + c(sin A – sin B)

= λ sin A (sin B – sin C) + λ sin B (sin C – sin A) + λ sin C (sin A – sin B)

= λ sin A sin B – λ sin A sin C + λ sin B sin C – λ sin B sin A) + λ sin C sin A – λ sin C sin B

= 0

As, LHS = RHS

Hence, Proved!

### Question 15: In △ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

LHS =

=

= λ2 sin A sin (B-C) + λ2 sin B sin (C-A) + λ2 sin C sin (A-B)

By using trigonometric identities,

sin(a-b) = sin a cos b – cos a sin b

= λ2 (sin A [sin B cos C – cos B sin C] + sin B [sin C cos A – cos C sin A] + sin C [sin A cos B – cos A sin B])

= λ2 (sin A sin B cos C – sin A cos B sin C + sin B sin C cos A – sin B cos C sin A + sin C sin A cos B – sin C cos A sin B)

= λ2 (0)

= 0

As, LHS = RHS

Hence, Proved!

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