Question

In a triangle ABC , the angle B is greater than angle A, if the values of the angle A and B satisfy the equation 3sinx 4$si{n}^3$x k = 0 , 0 < k < 1 , then value of C is

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

Answer 1

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

Given,

a $△ABC$ where $\angle B>\angle A$

$3\sin x-4{\sin }^{3}x-k=0$

$\because \sin 3x=3\sin x-4\sin x^{3}x$

Let,

$\sin 3A=\lambda$

$\sin 3B=\lambda$

$\sin 3A-\sin 3B=0$

$\because \sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

Now, $\frac{2\sin \left(\frac{3A-3B}{2}\right)\cdot 2\cos \left(\frac{3A+3B}{2}\right)}{A+B=60}=0$

$\because$ A+B = 60

Now, $In△ABC=\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow 60+\angle C={180}^{\circ }$

$\Rightarrow \angle C={120}^{\circ }$

$\text{option}C:\frac{2\pi }{3}$