Chapter 9.6, Problem 5WE

To determine

To find the measure of each numbers angle.

Answer to Problem 5WE

  $\angle 1={90}^0,\angle 2={90}^0,\angle 3={25}^0,\angle 4={65}^0,\angle 5={55}^0,\angle 6={125}^0,\angle 7={35}^0,\angle 8={60}^0,\angle 9={90}^0,\angle 10={60}^0$

Explanation of Solution

Given:

  $\overleftrightarrow{BZ}$ is a tangent to circle $O$ ; $AC$ is a diameter;

  $m\stackrel{⌢}{BC}=90$ ; $m\stackrel{⌢}{CD}=30$ ; $m\stackrel{⌢}{DE}=20$ .

  

Calculation:

Since $m\stackrel{⌢}{BC}=90$ ,

  $\Rightarrow \angle 1={90}^0$

  $\angle 2={90}^0$ (Angle in a semicircle is ${90}^0$ ).

Now, $m\stackrel{⌢}{EC}=m\stackrel{⌢}{CD}+m\stackrel{⌢}{ED}$

  $\Rightarrow m\stackrel{⌢}{EC}=50$

  $\Rightarrow \angle 3={25}^0$ (Angle at the center of a circle is twice the angle at the circumference.)

In $\Delta AEC$ ,

  $\angle 3+\angle 2+\angle 4={180}^0$ (Angle sum property of a triangle)

  $\Rightarrow \angle 4={65}^0$

  $\angle 5={55}^0$

Now , $\angle 6+\angle 5={180}^0$ (Linear Pair)

  $\Rightarrow \angle 6={125}^0$

Now, $\angle 5=\angle EFH={55}^0$ (Vertically opposite angles)

In $\Delta EFH$ ,

  $\angle EFH+\angle FEH+\angle 7={180}^0$ (Angle sum property of a triangle)

  $\Rightarrow \angle 7={35}^0$

Now in $\Delta CGF$ ,

  $\angle CGF+\angle 4+\angle 5={180}^0$ (Angle sum property of a triangle)

  $\Rightarrow \angle CGF={60}^0$

Now, $\angle 8=\angle CGF$ (Vertically opposite angles are equal)

  $\Rightarrow \angle 8={60}^0$

  $\Rightarrow \angle 9={90}^0$ (A tangent to a circle is perpendicular to the radius through the point of contact).

  $\Rightarrow \angle 10={60}^0$

Conclusion:

Therefore, $\angle 1={90}^0,\angle 2={90}^0,\angle 3={25}^0,\angle 4={65}^0,\angle 5={55}^0,\angle 6={125}^0,\angle 7={35}^0,\angle 8={60}^0,\angle 9={90}^0,\angle 10={60}^0.$