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
- A
\(\frac{1}{2}\)
- B
\(\frac{\sqrt{3}}{2}\)
- C
1
- D
\(\sqrt{3}\)
Answer & Solution
Correct Answer
(D)
\(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Since, \(A, B, C\) are in \(\mathrm{AP}\)
\[
\begin{aligned}
& \Rightarrow \quad 2 B=A+C \text { ie, } \angle B=60^{\circ} \\
& \therefore \frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =2 k(a \cos C+c \cos A) \\
& \quad\left[\begin{array}{l}
\text { using, } \\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}
\end{array}\right] \\
& =2 k(b) \\
& =2 \sin B
\end{aligned}
\]
\[
=\sqrt{3}
\]
\[
\text { [using, } b=a \cos C+c \cos A]
\]
\[
\begin{aligned}
& \Rightarrow \quad 2 B=A+C \text { ie, } \angle B=60^{\circ} \\
& \therefore \frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =2 k(a \cos C+c \cos A) \\
& \quad\left[\begin{array}{l}
\text { using, } \\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}
\end{array}\right] \\
& =2 k(b) \\
& =2 \sin B
\end{aligned}
\]
\[
=\sqrt{3}
\]
\[
\text { [using, } b=a \cos C+c \cos A]
\]
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