Question

In a $△ABC$, $\angle A$ is greater than $\angle B$. If the measures of $\angle A$ and $\angle B$ satisfy the equation $2\tan x-k(1+{\tan }^{2}x)=0$, where $kϵ(0,1)$, then the measure of the $\angle C$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{5\pi }{12}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (D)

$\frac{\pi }{2}$

$k=\frac{2\tan x}{1-{\tan }^{2}x}=\sin 2x$
Hence
$k=\sin 2A=\sin 2B$
Thus
$2A=\pi -2B$ .... the possibility of $A=B$ is ruled out since it is given $A>B$.
$A+B=\frac{\pi }{2}$
$\Rightarrow \pi -C=\frac{\pi }{2}$
$\Rightarrow C=\pi -\frac{\pi }{2}$
$\Rightarrow C=\frac{\pi }{2}$