In any triangle ABC,sin frac{A}{2} leq | vedclass

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(B) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Using the identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,

Vedclass · Accepted Answer

$\frac{a}{2 \sqrt{b c}}$

Vedclass · Answer

(B) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Using the identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we have $\sin \frac{A}{2} = \frac{a}{2R \cdot 2 \cos \frac{A}{2}} = \frac{a}{4R \cos \frac{A}{2}}$. Alternatively,using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s$ is the semi-perimeter. By the $AM$-$GM$ inequality,$\sqrt{(s-b)(s-c)} \leq \frac{(s-b)+(s-c)}{2} = \frac{2s-b-c}{2} = \frac{a}{2}$. Therefore,$\sin \frac{A}{2} = \frac{\sqrt{(s-b)(s-c)}}{\sqrt{bc}} \leq \frac{a}{2 \sqrt{bc}}$. Thus,$\sin \frac{A}{2} \leq \frac{a}{2 \sqrt{bc}}$.

In any triangle $ABC$,$\sin \frac{A}{2} \leq$

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