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 $\frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A$ is

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

Answer 1

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

$A,B$ and $C$ are in arithmetic progression

$\Rightarrow B=\frac{A+C}{2}$

$\Rightarrow 2B=A+C$ .......... $\left(1\right)$

$A+B+C={180}^o$ ...... $\left(2\right)$ [Property of triangle]

From $\left(1\right)$ and $\left(2\right)$, we get

$B={60}^o$

$\therefore \frac{a}{c}\sin 2C+\frac{c}{a}\sin 2A=\frac{\sin A}{\sin C}2\sin C\cos C+\frac{\sin C}{\sin A}2\sin A\cos A$

$=2\sin A\cos C+2\sin C\cos A$

$=2\sin (A+C)=2\sin B=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$

Hence, option D.