Question

If the angles $A,B$ and $C$ of a triangle are in an arithmetic progression and if $a,b$ and $c$ denote the lengths of the sides opposite to $A,B$ and $C$ respectively, then the value of the expression $\frac{a}{2}\sin 2C+\frac{c}{a}\sin 2A$ is

• A. $1$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is D $\sqrt{3}$

Angles $A,B,C$ are in A.P. hence $B=60°=\frac{\pi }{3}$
$\frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A=2\left(\frac{\sin C}{c}\right)a\cos C+2\left(\frac{\sin A}{a}\right)c\cos A$
$=2(\sin A\cos C+\cos A\sin C)$
$=2\sin (A+C)$
$=2\sin B$
$=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$