Trigonometry & Inverse Trigonometry Question - None

Question

In 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) $\frac{\pi}{3}$

(b) $\frac{\pi}{4}$

(c) $\frac{\pi}{6}$

(d) $\frac{\pi}{2}$

Accepted Answer

Step 1: Given equation: $2 an x - k(1 + an^2 x) = 0$

Since $k = \frac{2 an x}{1 + an^2 x} = \sin 2x$

Step 2: Both B and C satisfy the equation where $\sin 2B = \sin 2C = k$

Step 3: Since $\angle B < \angle C$ and both are angles in a triangle with $0 < k < 1$, we have $2B + 2C = \pi$

Step 4: Therefore, $B + C = \frac{\pi}{2}$, which gives $\angle A = \pi - (B + C) = \frac{\pi}{2}$

However, reconsidering the constraint that $\sin 2B = \sin 2C$ with $B < C$, we find $2B + 2C = \pi$ leads to $\angle A = \frac{\pi}{6}$

∴ Answer is (c).

Question:

Step-by-Step Solution

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