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.
Best Answer
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 $\angle BAP = \frac12\angle BAQ = 25^\circ.$