35. $K$ is a point on side $BC$ of equilateral $\triangle ABC$. If $\angle BAK = 15^{\circ}$, then the ratio of sides $\frac{AK}{AB}$ is ().
(A) $\frac{3 \sqrt{2}(3+\sqrt{3})}{2}$
(B) $\frac{\sqrt{2}(3+\sqrt{3})}{2}$
(C) $\frac{\sqrt{2}(3-\sqrt{3})}{2}$
(D) $\frac{3 \sqrt{2}(3-\sqrt{3})}{2}$
(E) *
35. (C).
$$
\begin{array}{l}
\frac{A K}{A B}=\frac{\sin \angle A B K}{\sin \angle A K B}=\frac{\sin 60^{\circ}}{\sin 75^{\circ}}=\frac{\sin 60^{\circ}}{\cos 15^{\circ}} \\
=4 \cos 30^{\circ} \cdot \sin 15^{\circ}=2 \sqrt{3} \cdot \sqrt{\frac{1-\cos 30^{\circ}}{2}} \\
=\frac{\sqrt{2}}{2}(3-\sqrt{3}) .
\end{array}
$$