[ Inscribed and circumscribed circles ] [ Properties and characteristics of a parallelogram ]

In parallelogram $P Q R S$, the bisector of the angle at vertex $P$, which is equal to $80^{\circ}$, intersects side $R S$ at point $L$. Find the radius of the circle that is tangent to segment $P Q$ and the rays $Q R$ and $P L$, given that $P Q=7$.
Let $M$ be the intersection point of the lines $P L$ and $Q R$. The desired circle is inscribed in the isosceles triangle $P Q M$.

## Solution

Let $M$ be the intersection point of the lines $P L$ and $Q R$, $O$ be the center of the given circle, $r$ be its radius, and $A$ be the point of tangency of the circle with the ray $P L$. Then,

$$
\angle Q M P = \angle S P M = \angle Q P M = 40^{\circ}.
$$

Therefore, triangle $P Q M$ is isosceles,

$$
P A = P Q \cos 40^{\circ} = 7 \cos 40^{\circ}, \quad r = O A = A P \operatorname{tg} 20^{\circ} = 7 \cos 40^{\circ} \operatorname{tg} 20^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-02.jpg?height=420&width=598&top_left_y=1185&top_left_x=730)

## Answer

$7 \cos 40^{\circ} \operatorname{tg} 20^{\circ}$.