Let sinA/sinB = sin(A-C)/sin(C-B),where A, B, C are angles of a triangle ABC. If the lengths of the sides

Question

Let \(\frac{sinA}{sinB}\) = \(\frac{sin(A-C)}{sin(C-B)}\), where A, B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :

sinA/sinB = sin(A-C)/sin(C-B)

(1) b² – a² = a² + c² 

(2) b²,c²,a² are in A.P. 

(3) c²,a²,b² are in A.P.

(4) a²,b²,c² are in A.P.

Answer 1

Option : (2). b²,c²,a² are in A.P. 

sinA/sinB = sin(A-C)/sin(C-B)

As A,B,C are angles of triangle 

A + B + C = π 

A = π – (B + C) 

So, 

sinA = sin(B + C) …(1) 

Similarly,

sinB = sin(A + C) …(2)

From (1) and (2),

\(\frac{sin(B+C)}{sin(A+C)}\) = \(\frac{sin(A-C)}{sin(C-B)}\)

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

sin²C-sin²B = sin²A-sin²C

{∵sin(x+y)sin(x-y) = sin²x-sin²y}

2sin²C = sin²A + sin²B

By sine rule,

2c² = a²+b²

⇒ b²,c² and a² are in A.P.