Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is C $\frac{\pi }{2}$

$k=\frac{2tanx}{1+ta{n}^{2}x}=sin2x\Rightarrow sin2C=sin2B$
But $\angle C>\angle B$
$2C=\pi -2B\Rightarrow B+C=\frac{\pi }{2}$
$\therefore \angle A=\frac{\pi }{2}$
option c