Question

If $\Delta =a^2-{(b-c)}^2$ and $\Delta$ is the area of the $\Delta ABC$

then $\tan A$ is equal to?

• A. $\frac{4}{15}$
• B. $\frac{8}{15}$
• C. $\frac{11}{15}$
• D. $\frac{13}{15}$

Answer 1

The correct option is B $\frac{8}{15}$

$\Delta =a^2-{(b-c)}^2\Rightarrow \Delta =(a-b+c)(a+b-c)$

$\Delta =2(s-b)2(s-c)$

$tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{(s-b)(s-c)}{\Delta }$(multiply and divide by (s-b)(s-c) inside the square root)

$\Rightarrow \frac{\Delta }{(s-b)(s-c)}=4$

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

and then $tanA=\frac{8}{15}$