Chapter 10.5, Problem 22PSB

To determine

To Find : $\text{The measures of}\angle A,\widehat{DE}\text{ and}\widehat{DF}$

Answer to Problem 22PSB

  $1.\angle A=42°2.\widehat{DE}=48°3.\widehat{DF}=132°$

Explanation of Solution

Given information :

1) $\overline{AC}$ and $\overline{EF}$ are diameters to outer and inner circles respectively.

2) $\overline{AB}$ is tangent to the inner circle.

3) $\overline{BC}=84°$

Formula Used:

1) Inscribed angle is half of the enclosed are

2) The angle between a secant and a tangent is equal to half the difference of the enclosed are

  $\begin{array}{l}\angle A=\frac{1}{2}\widehat{BC}\\ \angle A=\frac{1}{2}(84°)\\ \angle A=42°\end{array}$

Now

  $\begin{array}{l}\because \overline{EF}\text{ is diameter}\\ \widehat{\text{DE}}+\widehat{DF}=180°\mathrm{.......}(1)\end{array}$

Also

  $\begin{array}{l}\frac{1}{2}(\widehat{DF}-\widehat{DE})=\angle A\\ \frac{1}{2}(\widehat{DF}-\widehat{DE})=42°\\ \widehat{DF}-\widehat{DE}=84°\mathrm{.............}(2)\end{array}$

Solving (1) and (2)

  $\begin{array}{l}\widehat{DF}=132°\\ \widehat{DE}=48°\end{array}$