RD Chapter 10 Sine and Cosine Formulae and Their Applications Ex 10.1 Solutions
Question - 11 : - b sin B – c sin C = a sin (B – C)
Answer - 11 : -
By using the sine rule we know,
So, c = k sin C
Similarly, a = k sin A
And b = k sin B
We know,
Now let us consider LHS:
b sin B – c sin C
Substituting the values of b and c in the aboveequation, we get
k sin B sin B – k sin C sin C = k (sin2 B– sin2 C) ……….(i)
We know,
Sin2 B – sin2 C = sin(B + C) sin (B – C),
Substituting the above values in equation (i), we get
k (sin2 B – sin2 C) =k (sin (B + C) sin (B – C)) [since, π = A + B + C ⇒ B + C = π –A]
The above equation becomes,
= k (sin (π –A) sin (B – C)) [since, sin (π – θ) = sinθ]
= k (sin (A) sin (B – C))
From sine rule, a = k sin A, so the above equationbecomes,
= a sin (B – C)
= RHS
Hence proved.
Question - 12 : - a2 sin(B – C) = (b2 – c2) sin A
Answer - 12 : -
By using the sine rule we know,
So, c = k sin C
Similarly, a = k sin A
And b = k sin B
We know,
Now let us consider RHS:
(b2 – c2) sin A …
Substituting the values of b and c in the aboveequation, we get
(b2 – c2) sin A = [(k sinB)2 – ( k sin C)2] sin A
= k2 (sin2 B – sin2 C)sin A………. (i)
We know,
Sin2 B – sin2 C = sin(B + C) sin (B – C),
Substituting the above values in equation (i), we get
= k2 (sin (B + C) sin (B – C))sin A [since, π = A + B + C ⇒ B + C = π –A]
= k2 (sin (π –A) sin (B – C))sin A
= k2 (sin (A) sin (B – C)) sin A[since, sin (π – θ) = sin θ]
Rearranging the above equation we get
= (k sin (A))( sin (B – C)) (k sin A)
From sine rule, a = k sin A, so the above equationbecomes,
= a2 sin (B – C)
= RHS
Hence proved.
Question - 13 : - 
Answer - 13 : -
Question - 14 : - a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0
Answer - 14 : -
By using the sine rule we know,
a = k sin A, b = k sin B, c = k sin C
Let us consider LHS:
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)
Substituting the values of a, b, c from sine rule inabove equation, we get
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sinB) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B)
= k sin A sin B – k sin A sin C + k sin B sin C – ksin B sin A + k sin C sin A – k sin C sin B
Upon simplification, we get
= 0
= RHS
Hence proved.
Question - 15 : - 
Answer - 15 : -
Upon simplification we get,
= k2 [sin A sin (B – C) + sin B sin (C– A) + sin C sin (A – B)]
We know, sin (A – B) = sin A cos B – cos A sin B
Sin (B – C) = sin B cos C – cos B sin C
Sin (C – A) = sin C cos A – cos C sin A
So the above equation becomes,
= k2 [sin A (sin B cos C – cos B sinC) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)]
= k2 [sin A sin B cos C – sin A cos Bsin C + sin B sin C cos A – sin B cos C sin A + sin C sin A cos B – sin C cos Asin B)]
Upon simplification we get,
= 0
= RHS
Hence proved.