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)?
+9673 This is the diagram.
Note that \(\angle BPC = 180^\circ - 116^\circ = 64^\circ\).
Since BC = CP, \(\angle PBC = 64^\circ\).
Considering the interior angle sum of \(\triangle BPC\) gives \(\angle PCB = 52^\circ\).
Since AB = AC, \(\angle ACB = \angle ABC = \angle PBC = 64^\circ\).
Therefore, \(\angle ACP = \angle ACB - \angle PCB = 64^\circ - 52^\circ = 12^\circ\).
