What if you wanted to figure out the angle at which the sun's rays hit the earth? The sun’s rays hit the earth such that the tangent rays determine when daytime and night time are. The time and Earth’s rotation determine when certain locations have sun. If the arc that is exposed to sunlight is \begin{align*}178^\circ\end{align*}, what is the angle at which the sun’s rays hit the earth \begin{align*}(x^\circ)\end{align*}?

### Angles Outside a Circle

An angle is considered to be outside a circle if the vertex of the angle is outside the circle and the sides are tangents or secants. There are three types of angles that are outside a circle: an angle formed by two tangents, an angle formed by a tangent and a secant, and an angle formed by two secants. Just like an angle inside or on a circle, an angle outside a circle has a specific formula, involving the intercepted arcs.

#### Investigation: Find the Measure of an Angle outside a Circle

Tools Needed: pencil, paper, ruler, compass, protractor, colored pencils (optional)

- Draw three circles and label the centers \begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*}. In \begin{align*}\bigodot A\end{align*} draw two secant rays with the same endpoint, \begin{align*}\overrightarrow{DE}\end{align*} and \begin{align*}\overrightarrow{DF}\end{align*}. In \begin{align*}\bigodot B\end{align*}, draw two tangent rays with the same endpoint, \begin{align*}\overrightarrow{LM}\end{align*} and \begin{align*}\overrightarrow{LN}\end{align*}. In \begin{align*}\bigodot C\end{align*}, draw a tangent ray and a secant ray with the same endpoint, \begin{align*}\overrightarrow{QR}\end{align*} and \begin{align*}\overrightarrow{QS}\end{align*}. Label the points of intersection with the circles like they are in the pictures below.
- Draw in all the central angles: \begin{align*}\angle GAH, \angle EAF, \angle MBN, \angle RCT, \angle RCS\end{align*}. Then, find the measures of each of these angles using your protractor. Use color to differentiate.
- Find \begin{align*}m \angle EDF, m \angle MLN\end{align*}, and \begin{align*}m \angle RQS\end{align*}.
- Find \begin{align*}\frac{m \widehat{EF}-m \widehat{GH}}{2}, \frac{m \widehat{MPN}-m \widehat{MN}}{2}\end{align*}, and \begin{align*}\frac{m \widehat{RS}-m \widehat{RT}}{2}\end{align*}. What do you notice?

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

#### Solving for Unknown Values

1. Find the value of \begin{align*}x\end{align*}. You may assume lines that look tangent, are.

Set up an equation using the Outside Angle Theorem.

\begin{align*}\frac{(5x+10)^\circ-(3x+4)^\circ}{2} &= 30^\circ\\ (5x+10)^\circ-(3x+4)^\circ &= 60^\circ\\ 5x+10^\circ-3x-4^\circ &= 60^\circ\\ 2x+6^\circ &= 60^\circ\\ 2x &= 54^\circ\\ x &= 27^\circ\end{align*}

#### Measuring Angles

1. Find the value of \begin{align*}x\end{align*}.

\begin{align*}x=\frac{120^\circ - 32^\circ}{2}=\frac{88^\circ}{2}=44^\circ\end{align*}.

2. Find the value of \begin{align*}x\end{align*}.

First note that the missing arc by angle \begin{align*}x\end{align*} measures \begin{align*}32^\circ\end{align*} because the complete circle must make \begin{align*}360^\circ\end{align*}. Then, \begin{align*}x=\frac{141^\circ - 32^\circ}{2}=\frac{109^\circ}{2}=54.5^\circ\end{align*}.

#### Sun Problem Revisited

If \begin{align*}178^\circ\end{align*} of the Earth is exposed to the sun, then the angle at which the sun’s rays hit the Earth is \begin{align*}2^\circ\end{align*}. From the Outside Angle Theorem, these two angles are supplementary. From this, we also know that the other \begin{align*}182^\circ\end{align*} of the Earth is not exposed to sunlight and it is probably night time.

### Examples

Find the measure of \begin{align*}x\end{align*}.

For all of the problems below we can use the Outside Angle Theorem.

#### Example 1

\begin{align*}x=\frac{125^\circ-27^\circ}{2}=\frac{98^\circ}{2}=49^\circ\end{align*}

#### Example 2

\begin{align*}40^\circ\end{align*} is not the intercepted arc. Be careful! The intercepted arc is \begin{align*}120^\circ\end{align*}, \begin{align*}(360^\circ-200^\circ-40^\circ)\end{align*}. Therefore, \begin{align*}x=\frac{200^\circ-120^\circ}{2}=\frac{80^\circ}{2}=40^\circ\end{align*}.

#### Example 3

First, we need to find the other intercepted arc, \begin{align*}360^\circ-265^\circ=95^\circ\end{align*}. \begin{align*}x=\frac{265^\circ-95^\circ}{2}=\frac{170^\circ}{2}=85^\circ\end{align*}

### Interactive Practice

### Review

Find the value of the missing variable(s).

Solve for \begin{align*}x\end{align*}.

- Prove the Outside Angle Theorem

Given: Secant rays \begin{align*}\overrightarrow{AB}\end{align*} and \begin{align*}\overrightarrow{AC}\end{align*}

Prove: \begin{align*}m\angle a = \frac{1}{2} \left (m\widehat{BC}-m\widehat{DE} \right )\end{align*}

- Draw two secants that intersect:
- inside a circle.
- on a circle.
- outside a circle.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 9.8.