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:

The smallest angle lies opposite the shortest side, so use the cosine rule with the shortest side.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Sides are \(6+2\sqrt3\approx9.46\), \(4\sqrt3\approx6.93\) and \(2\sqrt6\approx4.90\). The smallest angle is opposite \(2\sqrt6\).

Step 2: Cosine rule
\(\cos\theta=\dfrac{(6+2\sqrt3)^2+48-24}{2(6+2\sqrt3)(4\sqrt3)}\).

Step 3: Simplify
Numerator \(=48+24\sqrt3+24=72+24\sqrt3\). Denominator \(=8\sqrt3(6+2\sqrt3)=48\sqrt3+48\).

Step 4: Reduce
\(\cos\theta=\dfrac{24(3+\sqrt3)}{48(\sqrt3+1)}=\dfrac{\sqrt3(\sqrt3+1)}{2(\sqrt3+1)}=\dfrac{\sqrt3}{2}\).

Step 5: Compare the options
So \(\theta=\frac{\pi}{6}\), the second option.

Final Answer:
The smallest angle is \(\frac{\pi}{6}\). \[ \boxed{\text{(B) }\dfrac{\pi}{6}} \]