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

### Question 1: If in â–³ABC, âˆ A=45Â°, âˆ B=60Â°, and âˆ C=75Â°, 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 (45Â°+30Â°) = sin(45Â°) cos(30Â°) + cos(45Â°) sin(30Â°)

sin 75Â° =

sin 75Â° =

Multiplying the denominator by 2âˆš2, we get

Hence, we get

a : b : c = 2 : âˆš6 : (âˆš3+1)

### Question 2: If in â–³ABC, âˆ C=105Â°, âˆ B=45Â°, and a=2, then find b.

Solution:

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

And as we know that,

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

A = Ï€-(B+C)

A = 180Â°-(45Â°+105Â°)

A = 30Â°

Now, according the sine rule

After rationalizing the denominator, we get

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

Solution:

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

Now, according the sine rule

Also,

### Question 4: In â–³ABC, 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

= Î»

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

= 2Î» sincoscos

= 2Î» sincoscos

= Î» sin(2cossin)

By using trigonometric formula,

2 sin a cos a = sin 2a

= Î» sin(sin)

= Î» sinsin(A+B)

= Î» sinsin(Ï€-C) (A+B+C = Ï€)

= Î» sinsin(C)

= (Î» sin C) sin

= c sin

As, LHS = RHS

Hence Proved!

### Question 6: In â–³ABC, 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 â–³ABC, 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 â–³ABC, 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 â–³ABC, 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 â–³ABC, 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 ((bÎ»)(cÎ»))

= Î»

= 2Î»

= Î»b2-Î»c2

= b(Î»b) – c(Î»c)

= 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

= Î»2 sin A(sin2 B – sin2 C)

By using trigonometric identities,

sin2 a – sin2 b = sin(a+b) sin (a-b)

= Î»2 sin A(sin(B+C) sin (B-C))

= Î»2 sin A(sin(\pi-A) sin (B-C))

= Î»2 sin A(sin(A) sin (B-C))

= Î»2 sin2 A sin (B-C)

= (Î» sin A)2 sin (B-C)

= a2 sin (B-C)

As, LHS = RHS

Hence, Proved!

### Question 13: In â–³ABC, 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)

= Î» sin A (sin B – sin C) + Î» sin B (sin C – sin A) + Î» sin C (sin A – sin B)

= Î» sin A sin B – Î» sin A sin C + Î» sin B sin C – Î» sin B sin A) + Î» sin C sin A – Î» sin C sin B

= 0

As, LHS = RHS

Hence, Proved!

### Question 15: In â–³ABC, prove the following:

Solution:

According to the sine rule

Considering LHS, we have

LHS =

=

= Î»2 sin A sin (B-C) + Î»2 sin B sin (C-A) + Î»2 sin C sin (A-B)

By using trigonometric identities,

sin(a-b) = sin a cos b – cos a sin b

= Î»2 (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])

= Î»2 (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)

= Î»2 (0)

= 0

As, LHS = RHS

Hence, Proved!

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