# 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.

Make $$CD=CA$$, and join $$PD$$. Then by SAS$$\triangle APC\cong\triangle DPC$$and$$\angle DPC=\angle APC=137^o$$Drawing $$PE\perp AC$$, since $$\angle EAP=30^o$$, then$$PE=PF=AF$$And joining $$FB$$, since $$FB\perp AP$$ then by SAS$$\triangle AFB\cong\triangle PFB$$and triangle $$ABP$$ is isosceles.
Therefore$$\angle ABP=34^o$$and$$\angle PBD=51^o-34^o=17^o$$And since$$\angle BDP=180^o-30^o=150^o$$then$$\angle BPD=180^o-17^o-150^o=13^o$$But$$\angle DPC=137^o$$Therefore$$\angle BPC=137^o+13^o=150^o$$
• Where did $FB \perp AP$ come from? Oct 10, 2018 at 12:02
• Fair question. I'm afraid I've assumed without proof that $BF$ is tangent to the circle at $F$, and so perpendicular to $PA$. This needs some work. Oct 10, 2018 at 17:54
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 $$\triangle BPC$$.