The smallest angle of the triangle whose sides are 6+12,48,24 is

cos⁡C=(6+12)2+48−242(6+12)48=36+12+1212+241248+212⋅48=72+121212⋅212+212+212=123(23+2)2412+48−123(23+2)24[12+2]=32⇒C=π6

The smallest angle of the triangle whose sides are 6+12,48,24 is

cos⁡C=(6+12)2+48−242(6+12)48=36+12+1212+241248+212⋅48=72+121212⋅212+212+212=123(23+2)2412+48−123(23+2)24[12+2]=32⇒C=π6

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

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$\frac{π}{3}$

$\frac{π}{6}$

$\frac{2 π}{3}$

$\frac{π}{4}$

answer is C.

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Detailed Solution

$\begin{aligned} \cos C = \frac{(6 + \sqrt{12})^{2} + 48 - 24}{2 (6 + \sqrt{12}) \sqrt{48}} = \frac{36 + 12 + 12 \sqrt{12} + 24}{12 \sqrt{48} + 2 \sqrt{12} \cdot \sqrt{48}} = \frac{72 + 12 \sqrt{12}}{12 \cdot 2 \sqrt{12} + 2 \sqrt{12} + 2 \sqrt{12}} \\ = \frac{12 \sqrt{3} (2 \sqrt{3} + 2)}{24 \sqrt{12} + 48} - \frac{12 \sqrt{3} (2 \sqrt{3} + 2)}{24 [\sqrt{12} + 2]} = \frac{\sqrt{3}}{2} \Rightarrow C = \frac{π}{6} \end{aligned}$