Search any question & find its solution
Question:
Answered & Verified by Expert
If in a $\triangle A B C, s(s-a)=(s-b)(s-c)$, then
Options:
Solution:
1480 Upvotes
Verified Answer
The correct answer is:
$\angle A=\frac{\pi}{2}$
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, $\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} \Rightarrow \cos ^2 \frac{A}{2}=\frac{s(s-a)}{b c}$
$$
\begin{array}{ll}
\text { and } \sin ^2(A / 2)=\frac{s(s-a)}{b c} \\
\therefore & \sin ^2(A / 2)=\cos ^2 \frac{A}{2} \\
\Rightarrow & \tan ^2(A / 2)=1 \Rightarrow \tan (A / 2)=1 \\
\Rightarrow & (A / 2)=\tan ^{-1} 1=(\pi / 4) \Rightarrow A=(2 \pi / 4)=\frac{\pi}{2} \\
\therefore & \angle A=\frac{\pi}{2}
\end{array}
$$
$$
\begin{aligned}
s(s-a) & =(s-b)(s-c) \\
\because \quad \sin \frac{A}{2} & =\sqrt{\frac{(s-b)(s-c)}{b c}}=\sqrt{\frac{s(s-a)}{b c}}
\end{aligned}
$$
Also, $\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}} \Rightarrow \cos ^2 \frac{A}{2}=\frac{s(s-a)}{b c}$
$$
\begin{array}{ll}
\text { and } \sin ^2(A / 2)=\frac{s(s-a)}{b c} \\
\therefore & \sin ^2(A / 2)=\cos ^2 \frac{A}{2} \\
\Rightarrow & \tan ^2(A / 2)=1 \Rightarrow \tan (A / 2)=1 \\
\Rightarrow & (A / 2)=\tan ^{-1} 1=(\pi / 4) \Rightarrow A=(2 \pi / 4)=\frac{\pi}{2} \\
\therefore & \angle A=\frac{\pi}{2}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.