[ Inscribed and Circumscribed Circles ] [ Inscribed Angle Subtended by a Diameter ]

A circle with radius 1 is circumscribed around triangle $A P K$. The extension of side $A P$ beyond vertex $P$ intercepts a segment $B K$ from the tangent to the circle at vertex $K$, and $B K$ is equal to 7. Find the area of triangle $A P K$, given that the angle $A B K$ is $\operatorname{arctg} \frac{2}{7}$.
Prove that triangle $A P K-$ is a right triangle.

## Solution

Let $O$ be the center of the given circle. Extend $K O$ to intersect the circle again at point $M$. From the right triangle $B K$, we find that

$$
\operatorname{tg} \angle M B K=\frac{K M}{B K}=\frac{2}{7}
$$

Therefore, $\angle M B K=\angle A B K$ and triangle $A P K-$ is a right triangle. Hence,

$$
\begin{gathered}
\angle A K P=\angle A B K, A P=A K \sin \angle A K P=2 \cdot \frac{2}{\sqrt{53}}=\frac{4}{\sqrt{53}} \\
P K=B K \sin \angle A B K=7 \cdot \frac{2}{\sqrt{53}}=\frac{14}{\sqrt{53}}
\end{gathered}
$$

Therefore,

$$
S_{\triangle \mathrm{APK}}=\frac{1}{2} A P \cdot P K=\frac{28}{53} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-06.jpg?height=446&width=878&top_left_y=563&top_left_x=590)

Answer

$\frac{28}{53}$.