Find the angle $x$ in this triangle 
This image was doing the rounds on a popular text messaging application, so I decided to give it a try.
From sine rule in $\triangle ABP$:
$$\frac{AB}{\sin(150^\circ)} = \frac{AP}{\sin(10^\circ} \\ \implies AP = 2AB \sin(10^\circ)$$
Applying sine rule again in $\triangle APC$:
$$\frac{AP}{\sin(60^\circ + x)} = \frac{AC}{\sin(x)}$$
Manipulating the equation and using some properties gives us
$$x = \arctan\left(\frac{\sqrt 3}{4\sin(10^\circ) - 1}\right)$$
This gives $x = -80^\circ$, but since it's an arctan, $x = 100^\circ$. Also, since $\sin(x) = \sin(\pi - x)$, $x = 80^\circ$ as well.
My question is: Is there a way to solve this problem that does not require a calculator? I tried to chase angles but that did not work out in this case. This solution requires computing $\sin(10^\circ)$ as well as the $\arctan$ of that expression, which needs a calculator.
 A: Let $D$ be the circumcenter of $\triangle APB$. Since $\angle BPA=150^\circ$, we have $\angle ADB = 360^\circ - 2\angle BPA = 60^\circ$, and since $DA=DB$ it follows that $\triangle ABD$ is equilateral. So, $AD=AB=AC$ and therefore $A$ is the circumcenter of $\triangle DBC$.

Now, $\angle PDB = 2\angle PAB = 40^\circ$ and $\angle CDB = \frac 12 \angle CAB = 40^\circ$. Hence $\angle PDB = \angle CDB$. It follows that $D,P,C$ are collinear. Now it is easy to find $\angle DPA = 90^\circ - \frac 12 \angle ADP = 90^\circ - \angle ABP = 80^\circ$. Thus $\angle APC = 180^\circ - \angle DPA = 100^\circ$.
A: Let $\angle ACP =z $. By trigonometric form of Ceva's theorem we have:
$$\frac {\sin60^\circ}{\sin20^\circ}\frac {\sin10^\circ}{\sin40^\circ}\frac {\sin(50^\circ-z)}{\sin z}=1.\tag1
$$
Further we have the following property for product of sines:
$$
\prod_{k=1}^{n-1}2\sin\frac{k\pi}n=n.
$$
Particularly for $n=9$ it gives
$$
(2^4\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ)^2=9
\implies\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac3{16}.\tag2
$$
Combining (1) and (2) we obtain:
$$
\frac {\sin z}{\sin(50^\circ-z)}=\frac{\sin^2 60^\circ\sin10^\circ\sin80^\circ}{\frac3{16}}
=4\sin10^\circ\cos10^\circ=\frac{\sin20^\circ}{\sin30^\circ}\implies z=20^\circ,
$$
and finally $$x=180^\circ-60^\circ-z=100^\circ.$$
