The rays $PX$ and $PY$ cut off arc $AB$ and $CD$ of a circle with radius $4$ . If the length of arc $CD$ is twice the length of the arc $AB$ and length of $CD$ is $\frac{4\pi}{5}$, find $\angle APB$ .
What I Tried :- Here is a diagram of my problem with my progress.
Now it is given that $CD = \frac{4\pi}{5}$ and $AB = \frac{2\pi}{5}$ .
Let $O$ be the center of the circle, I join $OA,OB,OC,OD,AB$ and $CD$ .
Now from the formula :- $$arc length = \frac{angle}{360^\circ} * 2\pi r$$
I find that $\angle AOB = 18^\circ$ and $\angle COD = 36^\circ$ . Now from the picture I get my remaining angles, after that I cannot proceed. I took $\angle OAC = x$ and $\angle DBO = y$, but I cannot find their values, can anyone help?
I also thought of using Power of a Point but I don't think I can use it here as you don't know any values of $PA,AC,PB,BD$ .
Best Answer
Since $ABDC$ is cyclic, $\angle BAC + \angle BDC = 180^\circ$.
That is, $x+81^\circ +72^\circ +y = 180^\circ$.
By angle sum of $\triangle PCD$, $\angle PCD + \angle CDP + \angle DPC = 180^\circ$.
That is, $x + 72^\circ + 72^\circ + y + \angle DPC = 180^\circ$.
Now solve for $\angle DPC$.