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In ∆Abc, Prove the Following: 2 ( B C Cos a + C a Cos B + a B Cos C ) = a 2 + B 2 + C 2

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प्रश्न

In ∆ABC, prove  the following: 

\[2 \left( bc \cos A + ca \cos B + ab \cos C \right) = a^2 + b^2 + c^2\]

 

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उत्तर

LHS = \[2 \left( bc \cos A + ca \cos B + ab \cos C \right)\]

On using the cosine law, we get: 

\[LHS = 2\left[ bc\left( \frac{b^2 + c^2 - a^2}{2bc} \right) + ca\left( \frac{a^2 + c^2 - b^2}{2ac} \right) + ab\left( \frac{a^2 + b^2 - c^2}{2ab} \right) \right]\]

\[= b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + a^2 + b^2 - c^2 \]

\[ = a^2 + b^2 + c^2 = RHS\]

Hence proved.

 

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Sine and Cosine Formulae and Their Applications
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