Question

The smallest angle of the triangle whose sides are $6 + \sqrt{12}, \sqrt{48}, \sqrt{24}$ is

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

Hint:

Smallest angle is opposite smallest side.

Answer 1

The Correct Option is C

Step 1: Simplify sides. \[ 6+\sqrt{12} = 6+2\sqrt{3}, \quad \sqrt{48}=4\sqrt{3}, \quad \sqrt{24}=2\sqrt{6} \] Smallest side = $2\sqrt{6}$
Step 2: Use cosine formula. Let angle opposite smallest side = $\theta$: \[ (2\sqrt{6})^2 = (6+2\sqrt{3})^2 + (4\sqrt{3})^2 - 2(6+2\sqrt{3})(4\sqrt{3})\cos\theta \]
Step 3: Solve. \[ 24 = (36 + 24\sqrt{3} + 12) + 48 - 2(24\sqrt{3} + 8\cdot3)\cos\theta \] \[ 24 = 96 + 24\sqrt{3} - (48\sqrt{3} + 48)\cos\theta \] \[ \Rightarrow \cos\theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{\pi}{4} \] Conclusion: Option (C)