[Law of Cosines] [Law of Sines]

Find the radius of the circumcircle of a triangle with sides 5 and 8 and an angle of $60^{\circ}$ between them.
Apply the formula $a=2 R \sin \alpha$.

## Solution

Let $ABC$ be the given triangle, $AB=5, AC=8, \angle BAC=60^{\circ}, R--$ the desired radius of the circumscribed circle. By the cosine rule,

$$
BC=\sqrt{AB^{2}+AC^{2}-2 AB \cdot AC \cos \angle BAC}=\sqrt{25+64-2 \cdot 5 \cdot 8 \cdot \frac{1}{2}}=\sqrt{49}=7
$$

Therefore,

$$
R=\frac{BC}{2 \sin \angle BAC}=\frac{7}{\sqrt{3}}=\frac{7 \sqrt{3}}{3}
$$

## Answer

$\frac{7 \sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-22.jpg?height=109&width=935&top_left_y=0&top_left_x=14)

Prove that the area of a trapezoid is equal to the product of the midline and the height.

## Hint

Draw a diagonal of the trapezoid.

## Solution

Let $CH$ be the height of trapezoid $ABCD$ with bases $AD$ and $BC, MN$ be the midline of the trapezoid. Draw diagonal $AC$ and height $AP$. Then

$$
\begin{gathered}
S(ABCD)=S_{\triangle ACD}+S_{\triangle ABC}=\frac{1}{2} AD \cdot CH+\frac{1}{2} BC \cdot AP= \\
=\frac{1}{2}(AD+BC) \cdot CH=MN \cdot CH
\end{gathered}
$$