Question

In a $\Delta ABC,\angle A<\angle B$. If $\sin A$ and $\sin B$ satisfy the equation $3\sin x-4{\sin }^{3}x-k=0$, where $0<K<1$, then find measure of $\angle C$.

• A. $\angle C=\frac{2\pi }{3}$
• B. $\angle C=\frac{3\pi }{4}$
• C. $\angle C=\frac{\pi }{4}$
• D. none of these

Answer 1

The correct option is A $\angle C=\frac{2\pi }{3}$

Given, in a $\Delta ABC,\angle A<\angle B$. If $\sin A$ and $\sin B$

satisfying the equation $3\sin x-4{\sin }^{3}x-k=0$, where $0<K<1$

$\Rightarrow \sin 3x-k=0$

$\Rightarrow \sin 3A=k$ and $\sin 3B=k$

$\Rightarrow 3A=3B$ or $\Rightarrow 3B=\pi -3A$

$\Rightarrow A+B=\frac{\pi }{3}$

$\therefore \angle C=\pi -(A+B)=\frac{2\pi }{3}$