Question

In a triangle \(ABC\), if \(\angle A=60^\circ\), then \((a+b+c)(b+c-a)=\)

• A. \(3bc\)
• B. \(2abc\)
• C. \(abc\)
• D. \(a+b+c\)

Hint:

Whenever a triangle has an angle of \(60^\circ\), the cosine rule simplifies to \[ a^2=b^2+c^2-bc. \] This identity is frequently useful in simplifying algebraic expressions involving the sides of a triangle.

Answer 1

The Correct Option is A

Step 1: Expand the given product using algebraic identities.
(a + b + c)(b + c - a) = [(b + c) + a][(b + c) - a] = (b + c)² - a² = b² + c² + 2bc - a².

Step 2: Invoke the Cosine Rule for angle A = 60°.
a² = b² + c² - 2bc cos 60°. Since cos 60° = 1/2, this becomes a² = b² + c² - bc.

Step 3: Substitute a² into the expanded expression.
b² + c² + 2bc - (b² + c² - bc) = b² + c² + 2bc - b² - c² + bc = 3bc.

Step 4: Final conclusion.
The expression simplifies to 3bc.