For any Δ ABC show that-
2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2
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Note: In any ΔABC we define ‘a’ as length of side opposite to ∠A , ‘b’ as length of side opposite to ∠B and ‘c’ as length of side opposite to ∠C .
Key point to solve the problem:
Idea of cosine formula in ΔABC
• Cos A =
• Cos B =
• Cos C =
As we have to prove:
2 (bc cos A + ca cos B + ab cos C) = a2 + b2 + c2
As LHS contain 2ca cos B, 2ab cos C and 2cb cos A ,which can be obtained from cosine formulae.
∴ From cosine formula we have:
Cos A =
⇒ 2bc cos A = …..eqn 1
Cos C =
⇒ 2ab cos C = …eqn 2
And, Cos B =
⇒ 2ac cos B = ……eqn 3
Adding eqn 1,2 and 3:-
2bc cos A + 2ab cos C + 2ac cos B = +
⇒ 2bc cos A + 2ab cos C + 2ac cos B =
⇒ 2(bc cos A + ab cos C + ac cos B) = …proved