%%

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 :

JEE · Math · previous-year question

  1. A.b2 $$-$$ a2 = a2 + c2
  2. B.b2, c2, a2 are in A.P.correct
  3. C.c2, a2, b2 are in A.P.
  4. D.a2, b2, c2 are in A.P.

Answer

B. b2, c2, a2 are in A.P.

Explanation

$${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$ 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 $$2{c^2} = {a^2} + {b^2}$$ $$\Rightarrow$$ b2, c2 and a2 are in A.P.

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions