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.

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 $$\angle PAB = 25^\circ$$
