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JEE Advanced · Mathematics · 17. Properties of Triangles

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

  1. A
    \(\frac{1}{2}\)
  2. B
    \(\frac{\sqrt{3}}{2}\)
  3. C
    1
  4. D
    \(\sqrt{3}\)
Verified Solution

Answer & Solution

Correct Answer

(D)
\(\sqrt{3}\)

Step-by-step Solution

Detailed explanation

Since, \(A, B, C\) are in \(\mathrm{AP}\)
\[
\begin{aligned}
& \Rightarrow \quad 2 B=A+C \text { ie, } \angle B=60^{\circ} \\
& \therefore \frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =2 k(a \cos C+c \cos A) \\
& \quad\left[\begin{array}{l}
\text { using, } \\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}
\end{array}\right] \\
& =2 k(b) \\
& =2 \sin B
\end{aligned}
\]

\[
=\sqrt{3}
\]
\[
\text { [using, } b=a \cos C+c \cos A]
\]
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