Question

In a triangle ABC, with usual notations, $(a + b + c)(a + b - c) = 3ab$, then $\angle C = $ ______.

• A. $\pi/2$
• B. $\pi/4$
• C. $\pi/3$
• D. $\pi/6$

Hint:

Any equation relating the squares of triangle sides ($a^2, b^2, c^2$) is almost guaranteed to be solved instantly by rearranging it into the numerator form of the Cosine Rule!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given an algebraic identity regarding the side lengths of a triangle. We must manipulate it into a standard trigonometric formula (the Cosine Rule) to solve for the specific angle $C$.

Step 2: Detailed Explanation:
The given equation is:
$(a + b + c)(a + b - c) = 3ab$
Recognize that the left side forms a classic "difference of squares" structure $(X + Y)(X - Y) = X^2 - Y^2$, where $X = (a+b)$ and $Y = c$:
$((a + b) + c)((a + b) - c) = 3ab$
$(a + b)^2 - c^2 = 3ab$
Expand the squared binomial:
$a^2 + b^2 + 2ab - c^2 = 3ab$
Rearrange the terms to isolate the core variables of the Cosine Rule ($a^2 + b^2 - c^2$):
$a^2 + b^2 - c^2 = 3ab - 2ab$
$a^2 + b^2 - c^2 = ab$
The fundamental Law of Cosines states that for any triangle:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Substitute our derived algebraic identity ($a^2 + b^2 - c^2 = ab$) directly into the numerator of the Cosine Rule:
$\cos C = \frac{ab}{2ab}$
The $ab$ terms perfectly cancel out (since side lengths cannot be zero):
$\cos C = \frac{1}{2}$
The principal angle in a triangle ($0 < C < 180^\circ$) whose cosine is exactly $\frac{1}{2}$ is $60^\circ$, which is $\pi/3$ in radians.

Step 3: Final Answer:
The angle C is $\pi/3$, matching option (c).