Question

In a triangle \(ABC\), \[ (b+c)\cos A+(c+a)\cos B+(a+b)\cos C= \]

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

Hint:

Projection formulas in a triangle are very useful: \[ a=b\cos C+c\cos B \] and cyclic variations help simplify expressions involving cosines quickly.

Answer 1

The Correct Option is C

Step 1: Use cosine rule formulas.
In a triangle, \[ \cos A=\frac{b^2+c^2-a^2}{2bc} \] \[ \cos B=\frac{c^2+a^2-b^2}{2ca} \] \[ \cos C=\frac{a^2+b^2-c^2}{2ab} \] Substitute these into the given expression: \[ (b+c)\cos A+(c+a)\cos B+(a+b)\cos C \]

Step 2: Simplify the first term.
\[ (b+c)\cos A = (b+c)\frac{b^2+c^2-a^2}{2bc} \] Using \[ a^2=b^2+c^2-2bc\cos A, \] we get \[ b^2+c^2-a^2=2bc\cos A \] Thus, \[ (b+c)\cos A = (b+c)\cdot\cos A \] Similarly for the remaining terms.
Instead of lengthy expansion, use the projection identity: \[ a=b\cos C+c\cos B \] \[ b=c\cos A+a\cos C \] \[ c=a\cos B+b\cos A \]

Step 3: Add the three identities.
Adding, \[ a+b+c = (b\cos C+c\cos B) + (c\cos A+a\cos C) + (a\cos B+b\cos A) \] Rearranging, \[ a+b+c = (b+c)\cos A + (c+a)\cos B + (a+b)\cos C \] Hence, \[ (b+c)\cos A+(c+a)\cos B+(a+b)\cos C = a+b+c \]

Step 4: Final conclusion.
Therefore, \[ \boxed{a+b+c} \] which corresponds to option (3).