Search any question & find its solution
Question:
Answered & Verified by Expert
If $s$ is the semi-perimeter of $\triangle A B C$ and if $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}$, then $\sum \sin ^2\left(\frac{A}{2}\right)=$
Options:
Solution:
1035 Upvotes
Verified Answer
The correct answer is:
$\frac{25}{33}$
Let $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}=k$
$\Rightarrow b+c-a=8 k, a+c-b=10 k$ and $a+b-c=12 k$
So, $a+b+c=30 k \Rightarrow s=15 k, a=11 k, b=10 k$ and $c=9 k$
$$
\text { Now, } \begin{aligned}
\Sigma \sin ^2\left(\frac{A}{2}\right)=\frac{(s-b)(s-c)}{b c}+\frac{(s-c)(s-a)}{c a} & \\
+ & \frac{(s-a)(s-b)}{a b}
\end{aligned}
$$
$\begin{aligned} & =\frac{30 k^2}{90 k^2}+\frac{24 k^2}{99 k^2}+\frac{20 k^2}{110 k^2} \\ & =\frac{1}{3}+\frac{8}{33}+\frac{2}{11}=\frac{11+8+6}{33}=\frac{25}{33} .\end{aligned}$
$\Rightarrow b+c-a=8 k, a+c-b=10 k$ and $a+b-c=12 k$
So, $a+b+c=30 k \Rightarrow s=15 k, a=11 k, b=10 k$ and $c=9 k$
$$
\text { Now, } \begin{aligned}
\Sigma \sin ^2\left(\frac{A}{2}\right)=\frac{(s-b)(s-c)}{b c}+\frac{(s-c)(s-a)}{c a} & \\
+ & \frac{(s-a)(s-b)}{a b}
\end{aligned}
$$
$\begin{aligned} & =\frac{30 k^2}{90 k^2}+\frac{24 k^2}{99 k^2}+\frac{20 k^2}{110 k^2} \\ & =\frac{1}{3}+\frac{8}{33}+\frac{2}{11}=\frac{11+8+6}{33}=\frac{25}{33} .\end{aligned}$
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.