Find angle inside isosceles triangle Please help. I have tried to find the answer but I couldn't. I have tried to draw a picture and measure the angle. It gives the answer $\theta = 25^\circ$ but I don't know how to do it.
Thank you

 A: Question is not incomplete.

If we partition $\angle PBC$ into $5^\circ$ and $30^\circ$ as in the figure, $PB$ becomes angle bisector of $\Delta ABR$. Now, notice that $\angle BRP = 60^\circ$. Also notice that since $\Delta BRC$ is isosceles as well, line segment $AR$ divides $\angle BAC$ equally. So $\angle RAC = 50^\circ$ and from there, $\angle ARP = 60^\circ$. Now notice that $PR$ is also angle bisector of $\Delta ABR$. Since two angle bisectors of $\Delta ABR$ intersects at $P$, third must pass through $P$ as well. Therefore $AP$ is the third angle bisector and $\angle PAR = \theta$. So we have $2\theta = 50 \implies \theta = 25^\circ$.
EDIT: Proof for the statement "If two angle bisectors intersects at point $P$, then third angle bisector also passes through $P$":

Assume we know that $BP$ and $CP$ are angle bisectors of $\Delta ABC$. Then let us draw perpendicular line segments $PD$, $PE$ and $PF$. Now notice that $\Delta BFP \cong \Delta BDP$ since they have the same angles and they share the hypotenus $BP$ (this means their hypotenus are the same). Therefore, we have $|PD| = |PF|$. By similar argument, $\Delta CDP \cong \Delta CEP$ so $|PD| = |PE|$. So we have $|PD| = |PE| = |PF|$. Now, since $|PE| = |PF|$ and $\Delta AFP$ and $\Delta AEP$ shares hypotenus, we have $|AF| = |AE|$ by Pythagoras Theorem. So, we also have $\Delta AFP \cong \Delta AEP$, which implies $\angle FAP = \angle PAE$.
I'm glad that you asked for a proof because I thought proof could come from inscribed circle but now I noticed that we can't know whether $P$ is the center of inscribed circle or not just by knowing two angle bisectors insersect at $P$, directly (it can be derived as in the following but not obvious in the beginning). When we use the argument above, we can conclude that $P$ is the center of inscribed circle by using the equation $|PD| = |PE| = |PF| = r$ where $r$ is the radius of inscribed circle. So my hint when you first asked for a proof was result of this argument but wasn't quite right, I admit that.
