Question

In triangle ABC, with usual notations, if \((a+b+c)(a+b-c) = ab\), then the measure of angle C is...

• A. \(\frac{π}{2}\)
• B. \(\frac{2π}{3}\)
• C. \(\frac{5π}{6}\)
• D. \(\frac{3π}{4}\)

Hint:

Expand the product and use the cosine rule.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The cosine rule states \(c^2 = a^2 + b^2 - 2ab\cos C\).

Step 2: Key Formula or Approach:
Expand \((a + b + c)(a + b - c) = (a + b)^2 - c^2\).

Step 3: Detailed Explanation:
\((a + b)^2 - c^2 = ab\), so \(a^2 + b^2 + 2ab - c^2 = ab\).
This gives \(a^2 + b^2 - c^2 = -ab\).
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{-ab}{2ab} = -\frac12 \]
So \(C = \frac{2\pi}{3}\).
The angle \(\frac{\pi}{2}\) would need \(\cos C = 0\), \(\frac{5\pi}{6}\) would give \(-\frac{\sqrt3}{2}\) and \(\frac{3\pi}{4}\) would give \(-\frac{1}{\sqrt2}\).

Final Answer:
\(C = \frac{2\pi}{3}\), option (B). \[ \boxed{\frac{2\pi}{3}} \]