Question

In a $\Delta$ ABC, $\angle B$ < $\angle C$ satisfy the euation $2tanx-k(1+ta{n}^{2}x)=0$, 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 $\frac{\pi }{2}$

ABC is a triangle.

$\angle B,\angle C$ satisfy $-k{\tan }^{2}x+2\tan x-k=0$

$\Rightarrow \tan B+\tan C=\frac{2}{k}$

$\Rightarrow \tan B\tan C=1$

$\Rightarrow \tan (B+C)=\frac{\tan B+\tan C}{1-\tan B\tan C}=\infty$

$\Rightarrow B+C=\frac{\pi }{2}$

$\therefore A=\pi -\frac{\pi }{2}=\frac{\pi }{2}$