Point within the interior of a given angle The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.

We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^2+PM^2+OK^2+KM^2\\=x^2+b^2+y^2+a^2.$$ I can't approach the problem further. How can we use the given angle $\alpha$? Thank you in advance!

Denote the mid-point of $OM$ be $N$. Then $N$ is the center of the circle passing through the points $O,P,M,K$.
We have:
$$\angle PNK = 2\alpha,\quad \angle PMK = \pi-\alpha$$
By Cosine Theorem,
$$PK^2 = a^2+b^2-2ab\cos \angle PMK = PN^2 + KN^2 - 2PN\cdot KN \cos 2\alpha$$
Writing $PN = KN = \frac12 OM = r$, we have:
$$a^2 + b^2 +2ab\cos \alpha = 2r^2(1-\cos 2\alpha)$$
or
$$r = \sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)}$$
To find $OM$ we just double this value:
$$OM = 2r = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)} = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(2 \sin^2 \alpha)} = \frac {\sqrt{a^2+b^2+2ab\cos \alpha}}{\sin \alpha}$$
