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In $\triangle \mathrm{ABC}$, if $\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{C}}{2}=\frac{\mathrm{b}}{\mathrm{s}}$, then $\sin \left(\frac{\mathrm{A}+\mathrm{C}}{3}\right)=$
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$\frac{1}{2}$
$\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{b}{s}$
$\Rightarrow \frac{\sin \frac{A}{2} \cdot \cos \frac{C}{2}+\cos \frac{A}{2} \cdot \sin \frac{C}{2}}{\cos \frac{A}{2} \cdot \cos \frac{C}{2}}=\frac{b}{s}$
$\begin{aligned} \Rightarrow & \frac{\sin \left(\frac{A+C}{2}\right)}{\sqrt{\frac{s(s-a)}{b c} \sqrt{\frac{s(s-c)}{a b}}}}=\frac{b}{s} \\ \Rightarrow & \frac{b \cdot \sin \left(\frac{A+C}{2}\right)}{s \sqrt{\frac{(s-a)(s-c)}{a c}}}=\frac{b}{s} \\ \Rightarrow & \frac{\sin \frac{A+C}{2}}{\sin \frac{B}{2}}=1 \Rightarrow \frac{\sin \frac{A+C}{2}}{\cos \frac{A+C}{2}}=1 \quad[\because A+B+C=\pi] \\ \Rightarrow & \tan \frac{A+C}{2}=1 \Rightarrow A+C=\frac{\pi}{2}\end{aligned}$
Now, $\sin \left(\frac{A+C}{3}\right)=\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$.
$\Rightarrow \frac{\sin \frac{A}{2} \cdot \cos \frac{C}{2}+\cos \frac{A}{2} \cdot \sin \frac{C}{2}}{\cos \frac{A}{2} \cdot \cos \frac{C}{2}}=\frac{b}{s}$
$\begin{aligned} \Rightarrow & \frac{\sin \left(\frac{A+C}{2}\right)}{\sqrt{\frac{s(s-a)}{b c} \sqrt{\frac{s(s-c)}{a b}}}}=\frac{b}{s} \\ \Rightarrow & \frac{b \cdot \sin \left(\frac{A+C}{2}\right)}{s \sqrt{\frac{(s-a)(s-c)}{a c}}}=\frac{b}{s} \\ \Rightarrow & \frac{\sin \frac{A+C}{2}}{\sin \frac{B}{2}}=1 \Rightarrow \frac{\sin \frac{A+C}{2}}{\cos \frac{A+C}{2}}=1 \quad[\because A+B+C=\pi] \\ \Rightarrow & \tan \frac{A+C}{2}=1 \Rightarrow A+C=\frac{\pi}{2}\end{aligned}$
Now, $\sin \left(\frac{A+C}{3}\right)=\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$.
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