Chapter 10.5, Problem 4PSA

a.

To determine

To find $m\angle K$

Answer to Problem 4PSA

  $m\angle K={42}^{\circ }$ .

Explanation of Solution

Given:

Given: $\stackrel{⌢}{HP}=120°$

  $\stackrel{⌢}{AM}=36°$

  

Calculation:

The measure of a secant-secant angle is one half the difference of the measures of the intercepted arcs.

  $\begin{array}{l}m\angle K=\frac{m\stackrel{⌢}{HP}-m\stackrel{⌢}{AM}}{2}\\ m\angle K=\frac{{120}^{\circ }-{36}^{\circ }}{2}\\ m\angle K={42}^{\circ }\end{array}$

b.

To determine

To find $m\angle T$

Answer to Problem 4PSA

  $m\angle T={50}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overline{TU}$ is tangent at U.

  $\stackrel{⌢}{RU}=160°$

  $\stackrel{⌢}{SU}=60°$

  

Calculation:

The measure of a secant-secant angle is one half the difference of the measures of the intercepted arcs.

  $\begin{array}{l}m\angle T=\frac{m\stackrel{⌢}{RU}-m\stackrel{⌢}{SU}}{2}\\ m\angle T=\frac{{160}^{\circ }-{60}^{\circ }}{2}\\ m\angle T={50}^{\circ }\end{array}$

c.

To determine

To find $m\angle X$

Answer to Problem 4PSA

  $m\angle x={40}^{\circ }$ .

Explanation of Solution

Given:

Given: W and R are point of contact.

  $\stackrel{⌢}{WR}=140°$

  

Calculation:

The measure of a secant-secant angle is one half the difference of the measures of the intercepted arcs.

  $\begin{array}{l}m\angle x=\frac{m\stackrel{⌢}{WYR}-m\stackrel{⌢}{WR}}{2}\\ m\angle x=\frac{{(360-140)}^{\circ }-{140}^{\circ }}{2}\\ m\angle x={40}^{\circ }\end{array}$