In A B C prove that A 2 A 2 A 2 =[A( A B C)]^2 a b c s - Vidyadip

Question

In $\triangle A B C$ prove that $\sin \frac{A}{2} \cdot \sin \frac{A}{2} \cdot \sin \frac{A}{2}=\frac{[A(\triangle A B C)]^2}{a b c s}$

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Answer

$\begin{aligned} & \text { LHS }=\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} \\ & =\sqrt{\frac{(s-b)(s-c)}{b c}} \cdot \sqrt{\frac{(s-a)(s-c)}{a c}} \cdot \sqrt{\frac{(s-a)(s-b)}{a b}}\end{aligned}$

$\begin{aligned} & =\sqrt{\frac{(s-a)^2(s-b)^2(s-c)^2}{a^2 b^2 c^2}} \\ & =\frac{(s-a)(s-b)(s-c)}{a b c}\end{aligned}$

$\begin{aligned} & =\frac{s(s-a)(s-b)(s-c)}{a b c s} \\ & =\frac{[\mathrm{A}(\triangle \mathrm{ABC})]^2}{a b c s} \ldots[\because \mathrm{A}(\triangle \mathrm{ABC})=\sqrt{s(s-a)(s-b)(s-c)]} \\ & =\text { RHS. }\end{aligned}$

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