JEE Main 2016 (21)

Geometry Level 4

$\cos(A-C)\cos(B)+\cos(2B)=0$ Then $a^{2},b^{2} \text{ and }c^{2}$ are in-

Clarification: $A,B \text{ and }C$ are angles of triangle, $a,b \text{ and }c$ are sides opposite to angles $A,B \text{ and } C$ respectively.

Arithmetic Progression Harmonic Progression Geometric Progression Arithmetico-Geometric Progression None of these

2 solutions

Aakash Khandelwal
Mar 23, 2016

Writing $cos(B)$ as $-cos(A+C)$ we get the above equation as $cos(A+C)cos(A-C)$ = $cos(2B)$ .

Use $cos(A+C)cos(A-C)$ = $1-sin^{2}A-sin^{2}C$

We get $sin^{2}A + sin^{2}C= 2sin^{2}B$ . Thereby using sine rule we get $a^{2} + c^{2} = 2b^{2}$ .

Ahmad Saad
Mar 22, 2016