Question

$ABC$ is triangle, if $\sin \left(\frac{A+B}{2}\right)=\frac{\sqrt{3}}{2}$, then the value of $\sin \frac{C}{2}$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C $\frac{1}{2}$

$\sin \left(\frac{A+B}{2}\right)=\frac{\sqrt{3}}{2}$
$\sin \left(\frac{A+B}{2}\right)=\sin {60}^{\circ }\Rightarrow \frac{A+B}{2}={60}^{\circ }$
$\angle A+\angle B={120}^{\circ }$
In a $\angle ABC$
$\angle A+\angle B+\angle C={180}^{\circ }$
$\angle C={60}^{\circ }$
$\sin \frac{C}{2}=\sin \frac{60}{2}=\sin {30}^{\circ }=\frac{1}{2}$.