If A, B and C are interior angle of a triangle ABC then show that $s i n (\frac{B + C}{2}) = c o s \frac{A}{2}$ .

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Solution

$∵ A, B, C$ are interior angles of a triangle ABC
$\Rightarrow A + B + C = 180^{\circ}$
$\Rightarrow A + B = 180^{\circ} - C$
$\Rightarrow \frac{A + B}{2} = 90^{\circ} - \frac{C}{2}$
$o r \frac{B + C}{2} = 90^{\circ} - \frac{A}{2} - (1)$
Taking in both sides
$\Rightarrow sin (\frac{B + C}{2}) = sin (90^{\circ} - \frac{A}{2})$
$\Rightarrow sin (\frac{B + C}{2}) = cos (\frac{A}{2})$

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Similar questions

A, B

C

A B C

sin (\frac{B + C}{2}) = cos \frac{A}{2} .

s i n \frac{(B + C)}{2} = c o s \frac{A}{2}

Δ

sin \frac{B + C}{2} = cos \frac{A}{2}

For a triangle ABC, show that sin $(\frac{B + C}{2})$ = cos $(\frac{A}{2})$ , where A, B and C are interior angles of $△$ ABC.

A, B, C

△ A B C

sin (\frac{B + C}{2}) = cos \frac{A}{2}

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