In triangle ABC, AB = AC. Point P lies on line AB such that CP = BC. If angle APC = 115 degrees, what is angle ACP(in degrees)?
So far I have figured out that the answer is not 65 or 70.
∠APC=115∘∠BPC=65∘∵CP=CB∴∠CBP=65∘∵AB=AC∠ACB=65∘
Then you consider △CPB to find angle PCB. The rest is trivial.