ABC is a triangle. If sin left(frac{A+B}{2}ri | vedclass

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Question

(C) Given that $\sin \left(\frac{A+B}{2}\right) = \frac{\sqrt{3}}{2}$.Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have $\frac{A+B}{2} = 60^{\circ}

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Given that $\sin \left(\frac{A+B}{2}\right) = \frac{\sqrt{3}}{2}$.Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have $\frac{A+B}{2} = 60^{\circ}$,which implies $A+B = 120^{\circ}$.In any triangle $ABC$,the sum of angles is $A+B+C = 180^{\circ}$.Substituting $A+B = 120^{\circ}$,we get $120^{\circ} + C = 180^{\circ}$,so $C = 60^{\circ}$.Now,we need to find $\sin \frac{C}{2} = \sin \frac{60^{\circ}}{2} = \sin 30^{\circ}$.Since $\sin 30^{\circ} = \frac{1}{2}$,the final value is $\frac{1}{2}$.

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

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