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

In $$\triangle$$ ABC, $$\angle$$ A + $$\angle$$ B + $$\angle$$ C = $$180^\circ$$

=> $$\angle A + \angle B = 180^\circ - \angle C$$

Dividing both sides by '2',

=> $$\frac{A+B}{2}=90^\circ-\frac{C}{2}$$ ----------(i)

Given : $$sin(\frac{A+B}{2})=\frac{\sqrt{3}}{2}$$

Substituting value from equation (i), we get :

=> $$sin(90^\circ-\frac{C}{2})=\frac{\sqrt{3}}{2}$$

Using, $$sin(90^\circ-\theta)=cos\theta$$

=> $$cos(\frac{C}{2})=\frac{\sqrt{3}}{2}$$

=> $$\frac{C}{2}=cos^{-1}(\frac{\sqrt{3}}{2})$$

=> $$\frac{C}{2}=30^\circ$$

=> $$\angle C = 30 \times 2 = 60^\circ$$

$$\therefore$$ $$sin\frac{C}{2}$$ = $$sin(\frac{60}{2})$$

= $$sin(30^\circ)=\frac{1}{2}$$

=> Ans - (C)