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 Concept:
We use algebraic expansion and the Cosine Rule for a triangle: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.

Step 2: Formula Application:
Expand the given equation: $((a+b) + c)((a+b) - c) = 3ab$. $(a+b)^2 - c^2 = 3ab$.

Step 3: Explanation:
$a^2 + b^2 + 2ab - c^2 = 3ab$ $a^2 + b^2 - c^2 = ab$. Now, substitute this into the Cosine Rule: $\cos C = \frac{ab}{2ab} = \frac{1}{2}$. Since $\cos C = 1/2$, $C = 60^\circ$ or $\pi/3$.

Step 4: Final Answer:
The angle $\angle C$ is $\pi/3$.