Chapter 9.6, Problem 7WE

To determine

To Find: The measure of each numbered angle.

Answer to Problem 7WE

The value measure of each numbered angle is

  $\begin{array}{l}\angle 1={90}^{\circ },\\ \angle 2={90}^{\circ },\\ \angle 3={25}^{\circ },\\ \angle 4={65}^{\circ },\\ \angle 5={55}^{\circ },\\ \angle 6={125}^{\circ },\\ \angle 7={35}^{\circ },\\ \angle 8={60}^{\circ },\\ \angle 9={90}^{\circ },\\ \angle 10={60}^{\circ }\end{array}$ .

Explanation of Solution

Given information:

  

Line $\overline{AC}$ is a diameter.

Line $\overleftrightarrow{BZ}$ is tangent to the center $\odot O$

  $\begin{array}{l}m\stackrel{⌢}{BC}=90°\hfill \\ m\stackrel{⌢}{CD}=30°\hfill \\ m\stackrel{⌢}{DE}=20°\hfill \end{array}$

An intercepted arc is equal to half of the measure of an inscribed angle.

So the value can be calculated as,

Line $\overline{AC}$ is a diameter.

Line $\overleftrightarrow{BZ}$ is tangent to the center $\odot O$

Also $AC\left|\right|BZ$ .

Hence, $\angle 1={90}^{\circ }$

  $\angle 2={90}^{\circ }\text{ (The angle in a semicircle is a right angle)}$

  $\begin{array}{l}\angle 3=\frac{1}{2}\left(m\stackrel{⌢}{CD}+m\stackrel{⌢}{DE}\right)\\ =\frac{1}{2}\left({30}^{\circ }+{20}^{\circ }\right)\\ ={25}^{\circ }\end{array}$

In $\Delta AEC$ ,

  $\begin{array}{l}\angle 2+\angle 3+\angle 4={180}^{\circ }\\ {90}^{\circ }+{25}^{\circ }+\angle 4={180}^{\circ }\\ \angle 4={65}^{\circ }\end{array}$

  $\begin{array}{l}\angle 7=\frac{1}{2}\left(m\stackrel{⌢}{AB}-m\stackrel{⌢}{DE}\right)\hfill \\ =\frac{1}{2}\left(90°–20°\right)\hfill \\ =\frac{1}{2}\times 70°\hfill \\ ={35}^{\circ }\hfill \end{array}$

In $\Delta EFG$ ,

  $\begin{array}{l}\angle FEG={90}^{\circ }\text{ (Linear pair with }\angle \text{2)}\\ \angle FEG+\angle 7+\angle EGF={180}^{\circ }\\ {90}^{\circ }+{35}^{\circ }+\angle EGF={180}^{\circ }\\ \angle EGF={55}^{\circ }\end{array}$

Now,

  $\begin{array}{l}\angle EGF=\angle 5\text{ (Vertically opp angle)}\\ \angle 5={55}^{\circ }\end{array}$

Now,

  $\begin{array}{l}\angle 6+\angle 5={180}^{\circ }\text{ (Linear Pair angles)}\\ \angle 6+{55}^{\circ }={180}^{\circ }\\ \angle 6={125}^{\circ }\end{array}$

In $\Delta HCD$ ,

  $\begin{array}{l}\angle CHD+\angle 4+\angle 5={180}^{\circ }\\ \angle CHD={180}^{\circ }-{65}^{\circ }-{55}^{\circ }\\ \angle CHD={60}^{\circ }\end{array}$

Now,

  $\begin{array}{l}\angle 8=\angle CHD\text{(Vertically opp angles)}\\ \angle 8={60}^{\circ }\end{array}$

  $\angle 9={90}^{\circ }(\text{Tangent}\text{is}\text{always}\text{perpendicular}\text{to}\text{radius}\text{at}\text{the}\text{point}\text{of}\text{contact})$

In $\Delta BOH$ ,

  $\begin{array}{l}\angle OBH+\angle 1+\angle 8={180}^{\circ }\\ \angle OBH={180}^{\circ }-{90}^{\circ }-{60}^{\circ }\\ \angle OBH={30}^{\circ }\end{array}$

Now,

  $\begin{array}{l}\angle 10+\angle OBH={90}^{\circ }\\ \angle 10={90}^{\circ }-{30}^{\circ }\\ \angle 10={60}^{\circ }\end{array}$

Therefore,

  $\begin{array}{l}\angle 1={90}^{\circ },\\ \angle 2={90}^{\circ },\\ \angle 3={25}^{\circ },\\ \angle 4={65}^{\circ },\\ \angle 5={55}^{\circ },\\ \angle 6={125}^{\circ },\\ \angle 7={35}^{\circ },\\ \angle 8={60}^{\circ },\\ \angle 9={90}^{\circ },\\ \angle 10={60}^{\circ }\end{array}$