What is the missing angle in the isosceles triangle? 
Here, $\Delta ABC$ is an isosceles triangle, where, $AB = AC$. $P$ is such a point interior to triangle $\Delta ABC$ so that some angles are formed inside, as shown in the figure.
$\angle BAP = ?$
 A: Construct the circle of center $A$ and radius $AB=AC$, and let $O$ be the point where line $CP$ intersects the circle again: we have $\angle BOC={1\over2}\angle BAC$ and $\angle OAB=2\angle BCO=60°$. It follows that $ABO$ is an equilateral triangle.
Construct now the circle of center $O$ and radius $OB=OA$ and observe that it passes through $P$ because $\angle BPA=150°$. We have then
$\angle BAP={1\over2}\angle BOP={1\over4}\angle BAC$. From this equality it follows that $\angle BAP={1\over3}\angle PAC=13°$.

A: 13
An isosceles triangle has two equal sides as well as two equal angles.
Thence triangles $\Delta ABP$ and $\Delta ACP$ have two sides in common.
Applying the Sine Rule to both gives
$$2\sin(2t-51)\sin(141-t) = \sin(t)$$
where $t$ is $\gamma_1$ in John's diagram. This has solution $t = 34$, leading to $\angle BAP = 13^\circ$.
[ $2\sin(17^\circ)\cos(17^\circ) = \sin(34^\circ)$ ]
I suspect there's an easier way to get there, but can't see it. (Probably involving $\gamma_1 = 2\beta_2$)
A: We have the following six linear equations in six variables: $\alpha,\alpha_1, \beta,\beta_1\beta_2,\gamma_1$


*

*$\alpha_1+39+\beta_1+\beta_2+\gamma_1+30=180$

*$\alpha_1+\beta_2+150=180$ 

*$39+\beta+\gamma_1=180$

*$\alpha+\beta_1+30=180$

*$\alpha+\beta+150=360$

*$\beta_1+\beta_2=\gamma_1+30$


Giving the following system
\begin{align}
\alpha && \alpha_1 && \beta && \beta_1 && \beta_2 && \gamma_1&&\\
0 && 1 && 0 && 1 && 1 && 1 &&\quad 111\\
0 && 1 && 0 && 0 && 1 && 0 && \quad30\\
0 && 0 && 1 && 0 && 0 && 1 &&\quad 141\\
1 && 0 && 0 && 1 && 0 && 0 &&\quad 150\\
1 && 0 && 1 && 0 && 0 && 0 && \quad210\\
0 && 0 && 0 && 1 && 1 && -1 &&\quad 30
\end{align}
Solving this by hand [ugh! wish I had a CAS] I get


*

*$\alpha=69+t$

*$\alpha_1=81-2t$

*$\beta=141-t$

*$\beta_1=81-t$

*$\beta_2=-51+2t$

*$\gamma_1=t$


In order for all six angle to be positive it must be the case that $\vert t-33\vert<7.5$.

