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

Answer 1

The correct option is C $\pi /2$

Given equation,
$2\tan x-k(1+{\tan }^{2}x)=0$
$k{\tan }^{2}x-2\tan x+k=0$
$tanC$ and $\tan B$ are the roots of the equation.
Thus, $\tan C+\tan B=\frac{2}{k}$ (Sum of the roots)
$\tan C\tan B=\frac{k}{k}=1$ (Product of the roots)
Now, $\tan (B+C)=\frac{\tan B+\tan C}{1-\tan A\tan B}$
$\tan (180-A)=\frac{\frac{2}{k}}{1-1}$ (Angle sum property)
$\tan A=\infty$
$\tan A=\tan 90$
$A={90}^{\circ }$