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

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

Answer 1

Answer: (D)

$\sqrt 3$

Answer 2

Since, A, B, C are in AP
$\Rightarrow 2 B = A + C \, i, e,., \, \angle B = 60^\circ$
$\therefore \frac{a}{c} (2 \, sin \, C \, cos \, C ) + \frac{c}{a}$ (2 sin A cos A)
\hspace26mm $\Bigg [ Using \, \frac{a}{ sin \, A } = \frac{b}{ sin \, B } = \frac{c}{ sin \, C } = \frac{1}{k} \Bigg]$
= 2 k (b)
= 2 sin B \hspace26mm [using b = a cos C + ccos A]
= $\sqrt 3$