Is there an elementary method of finding this missing angle? Let a point $P$ lie in a triangle $\triangle ABC$ such that $\angle BCP = \angle PCA = 13^\circ$,  $\angle CAP = 30^\circ$, and $\angle BAP = 73^\circ$.  Compute $\angle BPC$.

I have an ugly trig solution that looks something like this:

 Let $\angle PBC = \theta$.  It follows that $\angle PBA = 51-\theta$.  From trig Ceva, we see that: 
 
 $$\frac{\sin(30)}{\sin(73)}*\frac{\sin(51-\theta)}{\sin(\theta)}*\frac{\sin(13)}{\sin(13)} = 1$$
 
 Observe that $90-73=17$, and conveniently $17*3=51$.  This inspires the following manipulations:
 
 $$\frac{1}{2\sin(73)} * \frac{\sin(51-\theta)}{\sin(\theta)} = 1$$
$$\frac{1}{2\cos(17)} * \frac{\sin(51)\cos(\theta)-\cos(51)\sin(\theta)}{\sin(\theta)} = 1$$
$$\sin(51)\cos(\theta)-\cos(51)\sin(\theta)= 2\cos(17)\sin(\theta) $$
$$\sin(51)\cos(\theta) = 2\cos(17)\sin(\theta) + \cos(51)\sin(\theta)$$
$$\sin(51)\cos(\theta) = \sin(\theta)(2\cos(17) + \cos(51))$$
$$\tan(\theta) = \frac{\sin(51)}{2\cos(17) + \cos(51)}$$
 
 Proceeding with triple-angle formulae:
 
 $$\tan(\theta) = \frac{3\sin(17)-4\sin^3(17)}{2\cos(17) + 4\cos^3(17)-3\cos(17)}$$
$$\tan(\theta) = \frac{\sin(17)}{\cos(17)} * \frac{3-4\sin^2(17)}{4\cos^2(17)-1}$$
$$\tan(\theta) = \tan(17) * \frac{3-4(1-\cos^2(17))}{4\cos^2(17)-1}$$
$$\tan(\theta) = \tan(17) * \frac{4\cos^2(17)-1}{4\cos^2(17)-1}$$
$$\tan(\theta) = \tan(17)$$
 
 We conclude that $\theta = 17$ and $\boxed{\angle BPC = 150}$.

This is simply horrific.  Is there a more elegant method?  I notice that $73 = 13 + 60$, but I don't see where I would put an equilateral triangle.

Hint. Reflect $AC$ along $PC$ onto $BC$ to get the line $A'C$. To show that $\angle BPC=150^\circ$, it suffices to show that $\triangle BA'P$ is similar to $\triangle BPC$.
