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:

Whenever expressions involve terms like \(bc\cos A\), apply the cosine rule: \[ a^2=b^2+c^2-2bc\cos A \] to transform trigonometric expressions into algebraic ones.

Answer 1

The Correct Option is C

Step 1: Use the cosine rule.
In a triangle \(ABC\), by the cosine rule, \[ a^2=b^2+c^2-2bc\cos A \] Thus, \[ 2bc\cos A=b^2+c^2-a^2 \] Similarly, \[ 2ac\cos B=a^2+c^2-b^2 \] and \[ 2ab\cos C=a^2+b^2-c^2 \]

Step 2: Add the three expressions.
Adding, \[ 2bc\cos A+2ac\cos B+2ab\cos C \] \[ =(b^2+c^2-a^2) +(a^2+c^2-b^2) +(a^2+b^2-c^2) \] Now combine like terms: \[ =(-a^2+a^2+a^2) +(b^2-b^2+b^2) +(c^2+c^2-c^2) \] \[ =a^2+b^2+c^2 \] Therefore, \[ 2(bc\cos A+ac\cos B+ab\cos C) = a^2+b^2+c^2 \]

Step 3: Final conclusion.
Hence, \[ \boxed{a^2+b^2+c^2} \] which corresponds to option (3).