Question

Given that $\angle YPZ={110}^o$. The angle ∠XZY is degrees, where P is the center of the given circle.

Answer 1

Considering the arc minor ZY, $\angle ZXY=\frac{1}{2}\times \angle ZPY$
$\Rightarrow \angle ZXY=\frac{{110}^o}{2}={55}^o$

Given, XQ is perpendicular to ZY, and it passes through the center.
$\angle XQZ=\angle XQY={90}^o$
$\therefore Q$ is the midpoint, and ZQ = QY.

Also, QX is the common side of the $△ZXQ$ and $△YXQ.$
Hence, $△ZXQ\cong △YXQ$
(by SAS property)

$\therefore$ Corresponding angles must be equal, i.e.,
$\angle ZXQ=\angle YXQ=\frac{\angle YXZ}{2}$
$\Rightarrow \angle ZXQ=\frac{{55}^o}{2}={27.5}^o$

As $△ZXQ$ is a right triangle, $\angle XZQ={90}^o-{27.5}^o={62.5}^o$