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!
 A: 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:
$$OM = 2r = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(1-\cos2\alpha)} = 2\sqrt\frac {a^2+b^2+2ab\cos \alpha}{2(2 \sin^2 \alpha)} = \frac {\sqrt{a^2+b^2+2ab\cos \alpha}}{\sin \alpha}$$
A: Reflect $M$ across $OA$ and $OB$ and you get $A'$ resp. $B'$.
Let $t=AB$, then $A'B' = 2t$ and $OA'= OB' = OM = m$. Also $\angle A'OB' = 2\alpha$.
By the Law of cosine in triangle $ABM$ we have $$ t = \sqrt{a^2+b^2+2ab\cos \alpha}$$
and finally in isosceles triangle $A'B'O$ we have $$\boxed{m = {t\over \sin \alpha}}$$

A: Let $m=OM$, then $${a\over m} =  \sin \alpha _1$$ and $${b\over m} =  \sin \alpha _2$$
so $$\cos \alpha _1 = \sqrt{1-{a^2\over m^2}}\;\;\;\;\;\;\;\;\;\; \wedge \;\;\;\;\; \;\;\;\;\; \cos \alpha _2 = \sqrt{1-{b^2\over m^2}}$$
so $$\sin \alpha = {a\over m}\sqrt{1-{b^2\over m^2}}  + {b\over m} \sqrt{1-{a^2\over m^2}}$$
Now you have to find out $m$. :)
A: Let $d=OM$
$$ \arcsin(a/d)+ \arcsin(b/d) = \alpha$$
$$ a \sqrt{d^2-b^2} + b \sqrt{d^2-a^2} = d^2\sin \alpha$$
Let Mathematica solve fourth order equation for $d$
$$sal= \sin \alpha$$
{d -> 0.5` Sqrt[(4.` (1.` + 1.` b^2))/sal^2 + 
    2.` Sqrt[(4.` (1.` + 1.` b^2)^2)/sal^4 - (
      16.` (0.25` - 0.5` b^2 + 0.25` b^4 + 1.` b^2 sal^2))/sal^4]]}

A: A two-liner :

Since $OPMK$ is cyclic, $\angle PMK = \pi - \angle POK = \pi - \alpha$; by cosine-rule in $\triangle PMK$, $PK^2=a^2+b^2+2ab\cos \alpha$.
Now $OM$ is simply the diameter of the circle which can be found by sine-rule in either of $\triangle POK$ or $\triangle PMK$,
$$OM=2r=\frac{PK}{\sin \angle POK}=\frac{PK}{\sin \alpha}=\frac{\sqrt{a^2+b^2+2ab\cos \alpha}}{\sin \alpha}$$
A: Hint
Denote by $A$ the angle opposite to side with length $a$, similarly $B$.
You have
$$\sin A=\frac{a}{OM}, \, \sin B =\frac{b}{OM}$$ and
$$A+B= \alpha.$$
