12.177. Inside the given angle $\alpha$ there is a point at a distance $a$ from the vertex and at a distance $b$ from one side. Find the distance of this point from the other side.
Solution.

Let $OA$ be the distance from point $O$ to vertex $A$ of the given angle, $OB$ be the distance to one of the sides, $OC$ be the distance to the other side, $OA=a, OB=b, OC$, and $\angle BAC=\alpha$ (Fig. 12.47), $\angle OAB=x$.

In $\triangle ABO\left(\angle ABO=90^{\circ}\right): \sin x=\frac{OB}{OA}=\frac{b}{a} \cdot$ From $\triangle OCA\left(\angle OCA=90^{\circ}\right)$:

$OC=OA \sin \angle OAC=a \sin (\alpha-x)=a(\sin \alpha \cos x-\cos \alpha \sin x)=$ $=a\left(\sin \alpha \sqrt{1-\sin ^{2} x}-\cos \alpha \sin x\right)=a\left(\sin \alpha \sqrt{1-\frac{b^{2}}{a^{2}}}-\frac{b}{a} \cos \alpha\right)=$ $=\sin \alpha \sqrt{a^{2}-b^{2}}-b \cos \alpha$.

Answer: $\sin \alpha \sqrt{a^{2}-b^{2}}-b \cos \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0930.jpg?height=543&width=716&top_left_y=35&top_left_x=288)

Fig. 12.48