Question

In $\Delta ABC$, if $a=5$ and $\tan \frac{A-B}{2}=\frac{1}{4}\tan \frac{A+B}{2}$,
then $\sqrt{a^2-b^2}=$

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

$\because \tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}=\frac{a-b}{a+b}\tan \frac{A+B}{2}$
$\therefore \frac{a-b}{a+b}=\frac{1}{4}$
$\Rightarrow \frac{5-b}{5+b}=\frac{1}{4}$
$\Rightarrow 20-4b=5+b$
$\Rightarrow 5b=15\Rightarrow b=3$
$\therefore \sqrt{a^2-b^2}=\sqrt{25-9}=\sqrt{16}=4$