Question

In a triangle $ABC$, angle $B < $ angle $C$ and the values of B and C satisfy the equation $2 \tan x-k\left(1+\tan ^{2} x\right)=$ 0 where $(0< k <1)$. Then the measure of angle $A$ is :

• A. $\pi / 3$
• B. $2 \pi / 3$
• C. $\pi / 2$
• D. $3 \pi / 4$

Answer 1

Answer: (C)

$k =\frac{2 \tan x }{1+\tan ^{2} x }=\sin 2 x $
$\Rightarrow \sin 2 C =\sin 2 B$
But $\angle C >\angle B$
$2 C =\pi-2 B $
$\Rightarrow B + C =\frac{\pi}{2}$
$\therefore \angle A =\frac{\pi}{2}$