In a triangle ABC, angle B

Question

In a triangle ABC, angle B < angle C and the values of B and 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) π / 3 (B) 2 π / 3 (C) π / 2 (D) 3 π / 4

Tardigrade Admin · Accepted Answer

k =(2 tan x /1+ tan 2 x )= sin 2 x ⇒ sin 2 C = sin 2 B But angle C > angle B 2 C =π-2 B ⇒ B + C =(π/2) ∴ angle A =(π/2)

Q. In a triangle $A BC$ , angle $B <$ angle $C$ and the values of B and C satisfy the equation $2 tan x - k (1 + tan^{2} x) =$ 0 where $(0 < k < 1)$ . Then the measure of angle $A$ is :

$π /3$

$2 π /3$

$π /2$

$3 π /4$

$k = \frac{2 t a n x}{1 + t a n ^{2} x} = sin 2 x$
$\Rightarrow sin 2 C = sin 2 B$
But $∠ C > ∠ B$
$2 C = π - 2 B$
$\Rightarrow B + C = \frac{π}{2}$
$∴ ∠ A = \frac{π}{2}$

Q. In a triangle ABC, angle B< angle C and the values of B and C satisfy the equation 2tanx−k(1+tan2x)= 0 where (0<k<1). Then the measure of angle A is :

π/3

2π/3

π/2

3π/4