Question

Find the smallest angle of the triangle whose sides are \( 6 + \sqrt{12}, \sqrt{48}, \sqrt{24} \).

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

Hint:

Remember: The smallest angle in a triangle is opposite the shortest side. The Cosine Rule is helpful for finding the angles when you know the sides of the triangle.

Answer 1

The Correct Option is C

Given: Triangle side lengths are: \[ a = 6 + \sqrt{12}, \quad b = \sqrt{48}, \quad c = \sqrt{24} \] Step 1: Simplify Side Lengths Simplify the given lengths: \[ a = 6 + 2\sqrt{3}, \quad b = 4\sqrt{3}, \quad c = 2\sqrt{6} \] Step 2: Apply Cosine Rule Use the cosine rule: \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \] Substitute the simplified values: \[ \cos C = \frac{(6 + 2\sqrt{3})^2 + (4\sqrt{3})^2 - (2\sqrt{6})^2}{2(6 + 2\sqrt{3})(4\sqrt{3})} \] Simplify to determine angle \(C\).
Step 3: Conclusion The calculation yields the smallest angle \(C = \frac{\pi}{6}\). Answer: The correct answer is option (c): \( \frac{\pi}{6} \).