Question

If area of triangle $ABC$ is given by $△=a^2-(b-c{)}^2$ then $\tan \left(\frac{A}{2}\right)$ equals

• A. $-1$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{4}$

$△=(a+b-c)(a-b+c)$
$=(a+b+c-2c)(a+b+c-2b)$
$=4(s-c)(s-b)$
$\therefore \sqrt{s(s-a)(s-b)(s-c)}=4(s-c)(s-b)$
$\therefore \frac{1}{4}=\sqrt{\frac{(s-c)(s-b}{s(s-a)}}\Rightarrow \frac{1}{4}=\tan \left(\frac{A}{2}\right)$
Hence choice (c) is correct answer.