Let \[{{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}\], where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :
Solutions · 1
Official solution
\[{{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}\]
As A, B, C are angles of triangle.
A + B + C = \[\pi\]
A = \[\pi\] - (B + C) ...... (1)
Similarly sinB = sin(A + C) ..... (2)
From (1) and (2)
\[{{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}\]
\[\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)\]
\[{\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C\]
\[\because\] \[\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\} \]
\[2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B\]
By sine rule
2c2=a2+b2
\[\Rightarrow\] b2, c2 and a2 are in A.P.