Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. The area of $\Delta ABC$ is $a^2-{(b-c)}^2.$

Then $\tan A$ is equal to?

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{8}{15}$
• D. none of these

Answer 1

Answer: (C)

$\frac{8}{15}$

$\Delta =a^2-b^2-c^2+2bc$

$\Rightarrow \frac{\Delta }{2bc}=1-\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow \frac{\sin A}{4}=1-\cos A(\because \Delta =\frac{1}{2}bc\sin A)$

$\therefore (1-\frac{\sin A}{4}{)}^2+{\sin }^{2}A=1$

$\Rightarrow 1+\frac{{\sin }^{2}A}{16}-\frac{\sin A}{2}+{\sin }^{2}A=1$

$\Rightarrow \frac{17\sin A}{16}=\frac{1}{2}$

$\sin A=\frac{8}{17}$

$\therefore \tan A=\frac{\frac{8}{17}}{\sqrt{1-(\frac{8}{17}{)}^2}}=\frac{8}{15}$