Question

In $\Delta A B C$, 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)

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