Question

In a triangle $ABC$, angle $A$ is greater than $B$. If the measure of angles $A$ and $B$ satisfy the equation $3\sin x-4{\sin }^{3}x-k=0,0<k<1$, then the measure of angle $C$ is

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

Answer 1

Answer: (C)

$\frac{2\pi }{3}$

We have the expression $3\sin x-4{\sin }^{3}x-k=0\Rightarrow \sin 3x-k=0$ ...(1)

Since, $A$ and $B$ satisfy (1), therefore

$\sin 3A-k=0;\sin 3B-k=0\Rightarrow \sin 3A=\sin 3B$

Since $A>B$, therefore

$\sin 3A=\sin 3B\Rightarrow 3A={180}^0-3B\Rightarrow 3A+3B={180}^0\Rightarrow A+B={60}^0$

$\Rightarrow {180}^0-C={60}^0\Rightarrow C={120}^0=\frac{2\pi }{3}$