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 $\left(\frac{a}{c}\right)\sin 2C+\left(\frac{c}{a}\right)\sin 2A$ is

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

Answer 1

The correct option is D

$\sqrt{3}$

Explanation for Correct Option:

Step 1: Apply the Arithmetic formula

Since $A,B\mathrm{and}C$ are in arithmetic progression

$\Rightarrow B=\frac{A+C}{2}$

$\Rightarrow 2\mathrm{B}=\mathrm{A}+\mathrm{C}.....\left(\mathrm{i}\right)$

Sum of the angle of a triangle

$\mathrm{A}+\mathrm{B}+\mathrm{C}=180°.....\left(\mathrm{ii}\right)$

Step 1: On Solving equations (i) and (ii)

$\mathrm{B}=60°$

By sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$$\begin{gathered} \therefore \frac{\mathrm{a}}{\mathrm{c}}\sin 2\mathrm{C}+\frac{\mathrm{c}}{\mathrm{a}}\sin 2\mathrm{A}=\frac{\sin \mathrm{A}}{\sin \mathrm{C}}2\mathrm{sinC}\mathrm{cosC}+\frac{\sin \mathrm{C}}{\sin \mathrm{A}}2\mathrm{sinAcosA} \\ =2\mathrm{sinA}\mathrm{Sin}\mathrm{C}\frac{\cos \mathrm{C}}{\sin \mathrm{C}}+2\sin \mathrm{C}\mathrm{Sin}\mathrm{A}\frac{\cos \mathrm{A}}{\sin \mathrm{A}} \\ =2\sin \mathrm{A}\cos \mathrm{C}+2\mathrm{sinC}\cos \mathrm{A} \\ \Rightarrow 2\sin (\mathrm{A}+\mathrm{C}) \\ =2\mathrm{sinB} \\ =2\sin 60° \\ =2\times \frac{\sqrt{3}}{2} \\ =\sqrt{3} \end{gathered}$$

Hence option (4) is the Correct Option