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)?
+129845 A
P
B C
Since APC =115
Then angle BPC = 180 -115 = 65
And in triangle BPC, CP = BC
So angles BPC and PBC = 65
So angle PBC = angle ABC
And AB = AC
So angle ACB = angle ABC = 65
So angle BAC = 180 - 2(65) = 50 = angle PAC
So angle APC + angle PAC + angle ACP =180
So
115 + 50 + angle ACP =180
165 + angle ACP =180
angle ACP = 180 - 165 = 15°
