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If A,B,C are interior angles of …

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If A,B,C are interior angles of a triangles ABC then show that Sin(B+C)by 2 =cos A by 2

Posted by Md Danish 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

A, B, C, are interior angles of a {tex}\Delta {/tex}
{tex}\because A + B + C = 180 ^ { 0 }{/tex}
{tex}\Rightarrow B + C = 180 ^ { 0 } - A \Rightarrow \frac { B + C } { 2 } = 90 ^ { 0 } - \frac { A } { 2 }{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \sin \left( 90 ^ { \circ } - \frac { \mathrm { A } } { 2 } \right) \left[ \because \sin \left( 90 ^ { 0 } - \theta \right) = \cos \theta \right]{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \cos \frac { \mathrm { A } } { 2 } \text { proved }{/tex}
LHS = RHS

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