In $\triangle ABC$, prove that $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} = \dfrac{[A(\triangle ABC)]^2}{abcs}$.

In triangle ABC, prove that sin(A)/(2)sin(B)/(2)sin(C)/(2) = ([A(triangle ABC)]²)/(abcs).

Exercise 3.2 · Q47

\triangle ABC

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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$\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2=\sqrt{\dfrac{(s-b)(s-c)}{bc}}\cdot\sqrt{\dfrac{(s-a)(s-c)}{ac}}\cdot\sqrt{\dfrac{(s-a)(s-b)}{ab}}=\sqrt{\dfrac{(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2}}=\dfrac{(s-a)(s-b)(s-c)}{abc}$ (all factors positive, so the square root is the product itself). By Heron's Formula, $[A(\triangle ABC)]^2=s(s-a)(s-b)(s-c)\implies(s-a)(s-b)(s-c)=\dfrac{[A(\triangle ABC)]^2}{s}$ . Substituting: $\sin\dfrac A2\sin\df …

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Q.In $\triangle ABC$ , prove that $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} = \dfrac{[A(\triangle ABC)]^2}{abcs}$ .

Concept understanding — Half-Angle Formulae and Heron's Formula

Q.In △ABC\triangle ABC△ABC, prove that sin⁡A2sin⁡B2sin⁡C2=[A(△ABC)]2abcs\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} = \dfrac{[A(\triangle ABC)]^2}{abcs}sin2A​sin2B​sin2C​=abcs[A(△ABC)]2​.

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Q.In $\triangle ABC$ , prove that $\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2} = \dfrac{[A(\triangle ABC)]^2}{abcs}$ .

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