Question

In a triangle $ABC$, $2(bc \cos A + ac \cos B + ab \cos C) =$

• A. $a + b + c$
• B. $2(a + b + c)$
• C. $a^2 + b^2 + c^2$
• D. $2(a^2 + b^2 + c^2)$

Hint:

Combine cosine law identities to simplify cyclic trigonometric expressions in triangle geometry.

Answer 1

The Correct Option is C

Using the identity in a triangle: $a^2 = b^2 + c^2 - 2bc\cos A$
Similarly, $b^2 = a^2 + c^2 - 2ac\cos B$, and $c^2 = a^2 + b^2 - 2ab\cos C$
Adding the negatives: $-2(bc\cos A + ac\cos B + ab\cos C) = -(a^2 + b^2 + c^2 - a^2 - b^2 - c^2) = - (a^2 + b^2 + c^2 - a^2 - b^2 - c^2) = 0$
Hence, $2(bc\cos A + ac\cos B + ab\cos C) = a^2 + b^2 + c^2$