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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
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2823 Upvotes
Verified Answer
The correct answer is:
$\sqrt{3}$
$\sqrt{3}$
Since, $A, B, C$ are in $\mathrm{AP}$
$$
\begin{aligned}
& \Rightarrow \quad 2 B=A+C \text { ie, } \angle B=60^{\circ} \\
& \therefore \frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =2 k(a \cos C+c \cos A) \\
& \quad\left[\begin{array}{l}
\text { using, } \\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}
\end{array}\right] \\
& =2 k(b) \\
& =2 \sin B
\end{aligned}
$$
$$
=\sqrt{3}
$$
$$
\text { [using, } b=a \cos C+c \cos A]
$$
$$
\begin{aligned}
& \Rightarrow \quad 2 B=A+C \text { ie, } \angle B=60^{\circ} \\
& \therefore \frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =2 k(a \cos C+c \cos A) \\
& \quad\left[\begin{array}{l}
\text { using, } \\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}
\end{array}\right] \\
& =2 k(b) \\
& =2 \sin B
\end{aligned}
$$
$$
=\sqrt{3}
$$
$$
\text { [using, } b=a \cos C+c \cos A]
$$
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