[ Projections of the bases, sides, or vertices of a trapezoid ] [Isosceles, inscribed, and circumscribed trapezoids ]

The area of an isosceles trapezoid is $\sqrt{3}$. The angle between the diagonal and the base is $20^{\circ}$ more than the angle between the diagonal and the lateral side. Find the acute angle of the trapezoid if its diagonal is 2.
The height of an isosceles trapezoid, dropped from the vertex of the smaller base, divides the larger base into segments, one of which is equal to half the difference of the bases, and the other is half the sum.

## Solution

Let $AD$ be the larger base of the isosceles trapezoid $ABCD$. Then the angle $BAD$ is acute. If $CH$ is the height of the trapezoid, then

$$
DH = \frac{AD - BC}{2}, \quad AH = AD - DH = AD - \frac{AD - BC}{2} = \frac{AD + BC}{2},
$$

therefore

$$
S_{\mathrm{ABCD}} = \frac{AD + BC}{2} \cdot CH = AH \cdot CH = \sqrt{3}
$$

Let $\angle BAC = \alpha$. Then $\angle CAH = \alpha + 20^{\circ}$. From the right triangle $AHC$, we find that

$$
\begin{gathered}
AH = AC \cos \angle CAH = 2 \cos (\alpha + 20^{\circ}) \\
CH = AC \sin \angle CAH = 2 \sin (\alpha + 20^{\circ})
\end{gathered}
$$

therefore

$$
AH \cdot CH = 2 \cos (\alpha + 20^{\circ}) \cdot 2 \sin (\alpha + 20^{\circ}) = 2 \sin (2\alpha + 40^{\circ}) = \sqrt{3}.
$$

Thus, $2\alpha + 40^{\circ} = 60^{\circ}$ or $2\alpha + 40^{\circ} = 120^{\circ}$. From this, we find that $\alpha = 10^{\circ}$ or $\alpha = 40^{\circ}$, and $\angle BAD = 2\alpha + 20^{\circ} = 40^{\circ}$ or $\angle BAD = 100^{\circ}$. Since the angle $BAD$ is acute, only $40^{\circ}$ fits.

If $AD$ is the smaller base, then similarly we find that $\angle ABC = 80^{\circ}$.

## Answer

$40^{\circ}$ or $80^{\circ}$.

A tangent to a circle ($K$ is the point of tangency) is parallel to the chord $LM$. It is known that $LM = 6$, $KM = 5$. Find the radius of the circle.

## Hint

Apply the theorem about the angle between a tangent and a chord.

## Solution

Let point $B$ lie on the given tangent, with points $B$ and $L$ located on opposite sides of the line $KM$. By the theorem about the angle between a tangent and a chord,

$$
\angle KLM = \angle BKM = \angle KML,
$$

therefore, triangle $KLM$ is isosceles. If $KA$ is its height, then

$$
\begin{gathered}
MA = \frac{1}{2} ML = 3, \quad AK = \sqrt{KM^2 - AM^2} = \sqrt{25 - 9} = 4 \\
\sin \angle KML = \frac{AK}{KM} = \frac{4}{5}
\end{gathered}
$$

Let $R$ be the radius of the circle. Then

$$
R = \frac{KL}{2 \sin \angle KML} = \frac{KM}{2 \sin \angle KML} = \frac{5}{2 \cdot \frac{4}{5}} = \frac{25}{8}
$$

## Answer

$\frac{25}{8}$.