Question

In a triangle ABC, $\angle B<\angle C$ and the values of B & C satisfy the equation $2\tan x-k(1+{\tan }^{2}x)=0$, 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

Answer: (C)

$\frac{\pi }{2}$

Given $\angle B<\angle C$

$2\tan x-k(1+{\tan }^{2}x)=0$

$\Rightarrow 2\tan x=k(1+{\tan }^{2}x)$

$\Rightarrow \frac{2\tan x}{1+{\tan }^{2}x}=k$

$\Rightarrow \sin 2x=k$ $[\because \sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }]$

as $0<k<1$ let $k=\sin \alpha$

$\Rightarrow \sin 2x=\sin \alpha$

Both B and C satisfy the equation

So,$2B=2C=\alpha$ $\Rightarrow 2(B+C)=\pi$

but

$\angle B<\angle C$ $\Rightarrow 2(B+C)=\pi$

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

So

$\angle A=\pi -(\angle B+\angle C)$

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

$=\frac{\pi }{2}$

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