In triangle ABC, AB = AC. Point P lies on line AB such that CP = BC. If angle APC = \(116\) degrees, what is angle ACP(in degrees)?
First we know that ABC is an isosceles triangle. On line AB, there is a point P that makes CPB an isosceles triangle as well. Angle APC is 116. That means angle CPB is 180-116=64 degrees. Since triangle CPB is isosceles, angle PBC is also 64 degrees. That makes angle PCB 180-(64+64)=180-128=52 degrees. As you may recall, triangle ABC is isosceles, so angle ACP+52=64. Solving for the angle, we get ACP=64-52=12 degrees. (By the way, sorry there's no picture to show you my thinking.)