Angles on a point inside a triangle Let $ABC$ be an isosceles triangle with $AB=AC$  and $∠BAC =  100$. A point $P$
inside the triangle $ABC$ satisfies that $∠CBP=35$ and $∠PCB= 30$. Find the
measure, in degrees, of angle $∠BAP$. Attached is the figure of the triangle

I tried to Angle Chase but it seemed true for all values of $BAP$. I then tried using the sine law. In the triangle $PBC$, We have
$$\frac{PB \sin(35)}{\sin(30)}= PC$$
Trying it with triangles $APB$ ($x= BAP$) and the fact that they are isosceles
$$\frac{PC\sin(x+70)}{\sin(100-x)}=\frac{PB\sin(175-x)}{\sin (x)}$$
Which becomes, 
$$\frac{\sin(35)\sin(x+70)}{\sin(30)\sin(100-x)} = \frac{\sin(175-x)}{\sin(x)}$$
Where in I don't know how to solve it. Other methods are welcome.
 A: Let $M$ be the midpoint of $BC.$ Let $Q$ be the intersection of $CP$ with $AM.$
Note that $\angle ABC = \frac12(180^\circ - 100^\circ) = 40^\circ$ 
and therefore
$$ \angle PBA = \angle ABC - \angle PBC = 40^\circ - 35^\circ = 5^\circ.$$
Since $Q$ is on the perpendicular bisector of $BC,$ triangle $\triangle BQC$ 
is isoceles with $BQ = CQ.$
Moreover,
\begin{align}
\angle QBC &= \angle QCB = 30^\circ,\\
\angle PBQ &= \angle PBC - \angle QBC = 35^\circ - 30^\circ = 5^\circ,\\
\angle PQB &= \angle QCB + \angle QBC = 60^\circ,\\
\angle PQA &= \angle MQC = 90^\circ - \angle QCB = 60^\circ.\\
\end{align}
In summary, $\angle PBA = \angle PBQ$ and $\angle PQA = \angle PQB.$
That is, the rays $BP$ and $QP$ are the angle bisectors of angles 
$\angle ABQ$ and $\angle AQB$ of triangle $\triangle ABQ.$
The angle bisectors of a triangle are concurrent,
hence $AP$ is an angle bisector of $\angle BAQ.$
But $\angle BAQ = 50^\circ,$
so $\angle BAP = \frac12\angle BAQ = 25^\circ.$
A: Solution

Let $M$ be the midpoint of $BC$. $AM$ intersects $CP$ at $Q$. By the symmetry properties, we may have $$\angle QBC=\angle QCB=\angle PCB=30^o.$$
Thus, $$\angle PBQ=\angle PBC-\angle QBC=35^o-30^o=5^o.$$
But $$\angle ABP=\angle ABC-\angle PBC=\frac{1}{2}(180^o-\angle BAC)-\angle PBC=5^o.$$
Hence, $$\angle PBQ=\angle ABP,$$ namely, $BP$ bisects $\angle ABQ.$ Morover, since $$\angle AQP=\angle MQC=\angle MQB=90^o-\angle QCM=90^o-\angle PCB=60^o,$$ and $$\angle PQB=2\angle BCQ=2\angle BCP=60^o,$$hence $$\angle AQP=\angle PQB,$$namely, $ QP$ bisects $\angle AQB.$ To sum up, $P$ is the incenter of $\triangle ABQ$. Thus, $AP$ bisects $\angle BAQ$. It follows that $$\angle BAP=\frac{1}{2}\angle BAM= \frac{1}{4}\angle BAC=25^o.$$
A: Let $x=\angle APB$, $a=AB=AB$, $b=AP$. We have
$$\frac{\sin x}{a}=\frac{\sin 5^\circ }{b}$$
and 
$$\frac{\sin \left(245^{\circ} - x\right)}{a}=\frac{\sin 10^\circ }{b}$$
We have then:
$$\frac{\sin \left(245^{\circ} - x\right)}{\sin x}=\frac{\sin 10^\circ }{\sin 5^\circ }$$
Using the formula
$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\sin\beta\cos\alpha$$
we obtain:
$$\cot x = \frac{\sin 10^\circ }{\sin 245^\circ \sin 5^\circ } + \cot 245^\circ$$
Thus
$$x = \text{arccot}\left(\frac{\sin 10^\circ }{\sin 245^\circ  \sin 5^\circ } + \cot 245^\circ\right)$$
After wolframAlpha we have
$$x=150^{\circ}$$
Now it's easy to show, that $$\angle PAB = 25^\circ$$
