Chapter 9.7, Problem 5ST2

To determine

To find: The values of x, y .

Answer to Problem 5ST2

  $x=70°$ and $y=50°$

Explanation of Solution

Given:

The figure:

  

  $\overline{AB}$ , $\overline{AC}$ and $\overline{DE}$ are tangents.

Concept used:

The measure of the angle formed by two secants, two tangents or a secant and a tangent drawn from a point outside a circle is equal to half the difference of the measures of the intercepted arcs.

Calculation:

As $\overline{AB}$ and $\overline{AC}$ are tangents, therefore by the theorem, “The measure of the angle formed by\two tangents drawn from a point outside a circle is equal to half the difference of the measures of the intercepted arcs.”

The measure of angle x is

  $x°=\frac{1}{2}\left(m\stackrel{⌢}{BEC}-m\stackrel{⌢}{BC}\right)$

Now, $m\stackrel{⌢}{BEC}=m\stackrel{⌢}{BE}+m\stackrel{⌢}{EC}$

We know, $m\stackrel{⌢}{EC}=130°$

And

  $\begin{array}{l}m\stackrel{⌢}{BE}=360°-(m\stackrel{⌢}{BC}+m\stackrel{⌢}{EC})\\ m\stackrel{⌢}{BE}=360°-(110°+130°)\\ m\stackrel{⌢}{BE}=120°\end{array}$

Therefore,

  $\begin{array}{l}m\stackrel{⌢}{BEC}=m\stackrel{⌢}{BE}+m\stackrel{⌢}{EC}\\ m\stackrel{⌢}{BEC}=120°+130°\\ m\stackrel{⌢}{BEC}=250°\end{array}$

Thus,

  $\begin{array}{l}x°=\frac{1}{2}\left(m\stackrel{⌢}{BEC}-m\stackrel{⌢}{BC}\right)\\ x°=\frac{1}{2}\left(250°-110°\right)\\ x°=\frac{1}{2}\left(140°\right)\\ x°=70°\end{array}$

Now for y:

As $\overline{DE}$ is tangent and $\overline{DC}$ is a secant, then applying the theorem, “The measure of the angle formed by a secant and a tangent drawn from a point outside a circle is equal to half the difference of the measures of the intercepted arcs.”

We get, $\angle EDC=\frac{1}{2}\left(130°-y\right)$

Given $\angle EDC=40°$

Therefore,

  $\begin{array}{l}40°=\frac{1}{2}\left(130°-y\right)\\ 80°=130°-y\\ y=130°-80°\\ y=50°\end{array}$

Conclusion:

The value of x and y is: $x=70°$ and $y=50°$