Question

In the fig seg AB $\cong$ seg AC, seg PB $\cong$ seg PC, $\angle A={40}^{\circ },\angle P={140}^{\circ }$. Find $x$ and $y$.

Answer 1

In $△PBC$,
PB = PC
$\angle PBC=\angle PCB=x$ (Isosceles triangle property)
Sum of angle of $△PBC$,
$\angle PCB+\angle PBC+\angle BPC=180$
$x+x+140=180$x = 20^{\circ}$In$\triangle ABC$,AB=ACthus,$\angle ABC = \angle ACB = z$\left(Isoscelestriangleproperty\right)SumofanglesoftriangleABC,$\angle ABC + \angle ACB + \angle BAC = 180z + z + 40 = 1802z = 140z = 70^{\circ}$Now,$\angle ACP = \angle ACB - \angle PCBy = 70 - 20y = 50^{\circ}$