Question

In a triangle $ABC$ , if $\tan \frac{A}{2}=\frac{5}{6}$ and $\tan \frac{B}{2}=\frac{20}{37}$ , then $a+c$ is equal to

• A. $b$
• B. $2b$
• C. $3b$
• D. $4b$

Answer 1

Answer: (B)

Since, $\frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}$
$\Rightarrow$ $\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\tan \left(\frac{\pi }{2}-\frac{C}{2}\right)$
$\Rightarrow$ $\frac{\frac{5}{6}+\frac{20}{37}}{1-\frac{5}{6}\times \frac{20}{37}}=\cot \frac{C}{2}$
$\Rightarrow$ $\frac{305}{122}=\cot \frac{C}{2}$
$\Rightarrow$ $\tan \frac{C}{2}=\frac{2}{5}$
Now, $\tan \frac{A}{2}\tan \frac{C}{2}=\frac{5}{6}\times \frac{2}{5}$
$\Rightarrow$ $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{3}$
$\Rightarrow$ $\frac{s-b}{s}=\frac{1}{3}$
$\Rightarrow$ $3s-3b=s$
$\Rightarrow$ $2s=3b\Rightarrow a+b+c=3b$
$\Rightarrow$ $a+c=2b$