If the sides of a triangle are in the ratio $2: \sqrt{6}: 1+\sqrt{3}$, then what is the smallest angle of the triangle?

• A

  $75^{\circ}$
• B

  $60^{\circ}$
• Correct answer
  C

  $45^{\circ}$
• D

  $30^{\circ}$

Solution

$\begin{aligned} \cos A & =\frac{b^2+c^2-a^2}{2 b c} \\ & =\frac{(1+\sqrt{3})^2+(\sqrt{6})^2-(2)^2}{2(\sqrt{3}+1)(\sqrt{6})}\end{aligned}$ $\begin{aligned} & =\frac{1+3+2 \sqrt{3}+6-4}{2(\sqrt{3}+1)(\sqrt{6})} \\ & =\frac{2 \sqrt{3}+6}{2(\sqrt{3}+1)(\sqrt{6})}=\frac{2 \sqrt{3}(\sqrt{3}+1)}{2(\sqrt{3}+1)(\sqrt{6})}\end{aligned}$ $\begin{aligned} \cos \mathrm{A} & =\frac{1}{\sqrt{2}} \\ \cos \mathrm{~A} & =\cos 45^{\circ} \\ \mathrm{A} & =45^{\circ} .\end{aligned}$

Hence, the correct answer is option: 'C'.