Question

Find the value of $x$ if $\sin 2x=\sin {60}^{\circ }\cos {30}^{\circ }-\cos {60}^{\circ }\sin {30}^{\circ }$

• A. ${20}^{\circ }$
• B. ${15}^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${15}^{\circ }$

Given:

$\sin 2x=\sin {60}^{\circ }\cos {30}^{\circ }-\cos {60}^{\circ }\sin {30}^{\circ }$

$\sin 2x=\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$

$\therefore \sin 2x=\frac{1}{2}=\sin {30}^{\circ }$

$\Rightarrow 2x={30}^{\circ }\Rightarrow x={15}^{\circ }$