Prove that $\angle B C K=90^{\circ}+\angle B A K$ and apply the Law of Sines to triangles $A B K$ and $B K C$.

## Solution

Let $M$ be the point of intersection of the diagonals of quadrilateral $А В К С$. Denote $\angle B A K=\alpha$. Then

$$
\angle K M C=\angle A M B=60^{\circ}-\alpha, \angle B C K=180^{\circ}-\angle K M C-\angle M K C=90^{\circ}+\alpha .
$$

Applying the Law of Sines to triangles $А В К$ and $В С K$, we get:

$$
\begin{aligned}
& \frac{B K}{\sin \alpha}=\frac{A B}{\sin 30^{\circ}}=2 \sqrt{3} \\
& \frac{B K}{\cos \alpha}=\frac{B C}{\sin 60^{\circ}}=\frac{2}{\sqrt{3}}
\end{aligned}
$$

from which we find $\operatorname{tg} \alpha=\frac{1}{3}$. Therefore, $\cos \alpha=\frac{3}{\sqrt{10}}$. Consequently,

$$
B K=\frac{2}{\sqrt{3}} \cos \alpha=\sqrt{\frac{6}{5}}
$$

## Answer

The problem
[Law of Sines]

In a convex quadrilateral $A B K C$, side $A B$ is equal to $\sqrt{3}$, diagonal $B C$ is equal to 1, and the angles $A B C, B K A$, and $B K C$ are $120^{\circ}, 30^{\circ}$, and $60^{\circ}$ respectively. Find the side $B K$.