The angle at vertex $D$ of trapezoid $A B C D$ with bases $A D$ and $B C$ is $60^{\circ}$. Find the diagonals of the trapezoid, if $A D=$ $10, B C=3$ and $C D=4$.

#
Consider triangles $A C D$ and $B C D$.

## Solution

Using the Law of Cosines in triangles $A C D$ and $B C D$, we find that

$$
\begin{aligned}
& A C^{2}=A D^{2}+C D^{2}-2 A D \cdot C D \cos 60^{\circ}=100+16-40=76 \\
& B D^{2}=B C^{2}+C D^{2}-2 B C \cdot C D \cos 120^{\circ}=9+16+12=37
\end{aligned}
$$

Therefore, $A C=2 \sqrt{19}, B D=\sqrt{37}$.

## Answer

$2 \sqrt{19} ; \sqrt{37}$.