Question

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

• A. -1
• B. 0
• C. 1/4
• D. 1/2

Answer 1

The correct option is C 1/4

Given $\Delta =a^2-(b-c{)}^2=(a+b-c\left)\right(a-b+c)=(a+b+c-2c\left)\right(a+b+c-2b)=(2s-2c\left)\right(2s-2b)$

$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=4(s-b)(s-c)$

$\Rightarrow \frac{1}{4}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=tan\left(\frac{A}{2}\right)$

$\therefore tan\left(\frac{A}{2}\right)=\frac{1}{4}$