Chapter 10.5, Problem 14PSA

To determine

To Find : $\text{Measure of}\stackrel{⌢}{WZ}$

Answer to Problem 14PSA

  ${110}^{\circ }$

Explanation of Solution

Given information :

1) $\overline{WY}$ is a diameter

2) $\stackrel{⌢}{WX}={50}^{\circ }$

3) $\angle XPY={120}^{\circ }$

Formula Used:

1) $\frac{1}{2}\left(s_1+s_2\right)=\theta$ (angle between chords)

2) Angle subtended by semi-circle is ${180}^{\circ }$

Calculation :

  $\begin{array}{l}\angle XPW+\angle XPY={180}^{\circ }\\ \angle XPW+{120}^{\circ }={180}^{\circ }\\ \angle XPW={60}^{\circ }\end{array}$

And

  $\begin{array}{l}\frac{1}{2}\left(\stackrel{⌢}{WX}+\stackrel{⌢}{YZ}\right)=\angle XPQ\\ \frac{1}{2}\left({50}^{\circ }+\stackrel{⌢}{YZ}\right)={60}^{\circ }\\ {50}^{\circ }+\stackrel{⌢}{YZ}={120}^{\circ }\\ \stackrel{⌢}{YZ}={70}^{\circ }\end{array}$

Now, $\overline{WY}$ is a diameter

  $\begin{array}{l}\stackrel{⌢}{WZ}+\stackrel{⌢}{YZ}={180}^{\circ }\\ \stackrel{⌢}{WZ}+{70}^{\circ }={180}^{\circ }\\ \stackrel{⌢}{WZ}={110}^{\circ }\end{array}$