Question

If the angles A, B and C of a triangle are in an AP and if a, b and c denote the length 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 $\sqrt{3}$

By the sine rule, we can replace the ratios $\frac{a}{c}$ and

$\frac{c}{a}$ by the ratio of the sines of opposite angles.

Then the given expression becomes simply
$2(\sin A\cos C+\cos A\sin C)=2\sin (A+C)=2\sin B$

Bit as the angles are in A.P $\Rightarrow \angle B=\frac{\pi }{3}$

So the given expression is
$2\sin \left(\frac{\pi }{3}\right)=\sqrt{3}$