Question

In any triangle $ ABC $, evaluate: $$ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} $$

• A. \( a^2 + b^2 + c^2 \)
• B. \( \frac{a^2 + b^2 + c^2}{2abc} \)
• C. \( \frac{2abc}{a^2 + b^2 + c^2} \)
• D. \( a + b + c \)

Hint:

Apply cosine law and normalize each cosine term by its corresponding side to derive identities in triangles.

Answer 1

The Correct Option is B

Use the identity: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \Rightarrow \frac{\cos A}{a} = \frac{b^2 + c^2 - a^2}{2abc} \] Similarly: \[ \frac{\cos B}{b} = \frac{c^2 + a^2 - b^2}{2abc},\quad \frac{\cos C}{c} = \frac{a^2 + b^2 - c^2}{2abc} \] Now add all: \[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{(b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)}{2abc} \] Simplify numerator: \[ = b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2 = a^2 + b^2 + c^2 \] So: \[ \frac{a^2 + b^2 + c^2}{2abc} \]