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 expression.
Let \[ E=(a+b+c)(b+c-a) \] Using the identity \[ (x+y)(y-x)=y^2-x^2, \] we get \[ E=(b+c+a)(b+c-a) \] \[ E=(b+c)^2-a^2 \] \[ E=b^2+c^2+2bc-a^2 \]

Step 2: Apply the Cosine Rule.
Since \[ \angle A=60^\circ, \] by the cosine rule, \[ a^2=b^2+c^2-2bc\cos 60^\circ \] Since \[ \cos 60^\circ=\frac12, \] \[ a^2=b^2+c^2-bc \]

Step 3: Substitute the value of \(a^2\).
Substituting into \(E\), \[ E=b^2+c^2+2bc-(b^2+c^2-bc) \] \[ E=b^2+c^2+2bc-b^2-c^2+bc \] \[ E=3bc \]

Step 4: Final conclusion.
Therefore, \[ \boxed{3bc} \]