If in a $\triangle A B C, s(s-a)=(s-b)(s-c)$, then
If in a $\triangle A B C, s(s-a)=(s-b)(s-c)$, then
- $\angle A=\frac{\pi}{4}$
- $\angle B=\frac{\pi}{3}$
- $\angle A=\frac{\pi}{2}$
- $\angle B=\frac{\pi}{2}$
Solution
Given, $\triangle A B C$ and
$
\begin{aligned}
s(s-a) & =(s-b)(s-c) \\
\because \quad \sin \frac{A}{2} & =\sqrt{\frac{(s-b)(s-c)}{b c}}=\sqrt{\frac{s(s-a)}{b c}}
\end{aligned}
$
Also, $\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} \Rightarrow \cos ^2 \frac{A}{2}=\frac{s(s-a)}{b c}$
$
\begin{array}{ll}
\text { and } \sin ^2(A / 2)=\frac{s(s-a)}{b c} \\
\therefore & \sin ^2(A / 2)=\cos ^2 \frac{A}{2} \\
\Rightarrow & \tan ^2(A / 2)=1 \Rightarrow \tan (A / 2)=1 \\
\Rightarrow & (A / 2)=\tan ^{-1} 1=(\pi / 4) \Rightarrow A=(2 \pi / 4)=\frac{\pi}{2} \\
\therefore & \angle A=\frac{\pi}{2}
\end{array}
$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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