Question

In the figure, $\angle$QXR$={25}^o$, $\angle$QRX$={33}^o$. Find $\angle$XYZ and $\angle$PZQ.

Answer 1

Given that:

$\angle QXR={25}^{\circ },\angle QRX={33}^{\circ }$

To find:

$\angle XYZ=?$

$\angle PZQ=?$

Solution:

In $△XQR$

$\angle XQR+\angle QXR+\angle QRX={180}^{\circ }$

or, $\angle XQR+{25}^{\circ }+{33}^{\circ }={180}^{\circ }$

or, $\angle XQR={122}^{\circ }$

Now,

$\angle PQX+\angle XQR={180}^{\circ }$ (Linear pair)

or, $\angle PQX={180}^{\circ }-{122}^{\circ }={58}^{\circ }$

$\angle PQX=\angle XYZ$ (Angles in the same segment are equal.)

or, $\angle XYZ=\angle PQX={58}^{\circ }$

Now,

$\angle PZX=2\angle XYZ$ (Angle subtended by an arc at the centre is twice of the angle subtended by it at any point on the circle.)

or, $\angle PZX=2\times {58}^{\circ }={116}^{\circ }$

$\angle PZX+\angle PZQ={180}^{\circ }$ (Linear pair)

or, $\angle PZQ={180}^{\circ }-{116}^{\circ }={64}^{\circ }$