10.243. Inside an angle of $60^{\circ}$, there is a point located at distances of $\sqrt{7}$ and $2 \sqrt{7}$ cm from the sides of the angle. Find the distance from this point to the vertex of the angle.
Solution.

Let point $O$ be the vertex of the angle ($\angle O=60^{\circ}$), $AB$ and $AC$ be the distances from point $A$ to the sides, $AB=\sqrt{7}$ cm, $AC=2\sqrt{7}$ cm (Fig. 10.50). $D$ is the intersection point of line $AB$ and ray $OC$, $\angle D=30^{\circ}$. From $\angle ACD$ ($\angle ACD=90^{\circ}$): $AD=2AC=4\sqrt{7}$ cm. $BD=AB+AD=5\sqrt{7}$ cm.

From $\triangle OBD$ ($\angle OBD=90^{\circ}$): $OB=BD \operatorname{tg} \angle D=\frac{5\sqrt{7}}{\sqrt{3}}$ cm. From $\triangle OBA$ ($\angle OBA=90^{\circ}$): $OA=\sqrt{OB^2+AB^2}=\frac{14\sqrt{3}}{3}$ cm.

Answer: $\frac{14\sqrt{3}}{3}$ cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0743.jpg?height=409&width=840&top_left_y=58&top_left_x=258)

Fig. 10.51