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# 9.5: Angles of Chords, Secants, and Tangents

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the measures of angles formed by chords, secants, and tangents.

## Review Queue

1. What is $m\angle OML$ and $m\angle OPL$? How do you know?
2. Find $m\angle MLP$.
3. Find $m\widehat{MNP}$.

Know What? The sun’s rays hit the Earth such that the tangent rays determine when daytime and night time are. If the arc that is exposed to sunlight is $178^\circ$, what is the angle at which the sun’s rays hit the earth $(x^\circ)$?

## Angle on a Circle

When an angle is on a circle, the vertex is on the edge of the circle. One type of angle on a circle is the inscribed angle, from the previous section. Another type of angle on a circle is one formed by a tangent and a chord.

Investigation 9-6: The Measure of an Angle formed by a Tangent and a Chord

Tools Needed: pencil, paper, ruler, compass, protractor

1. Draw $\bigodot A$ with chord $\overline{BC}$ and tangent line $\overleftrightarrow{ED}$ with point of tangency $C$.
2. Draw in central angle $\angle CAB$. Find $m\angle CAB$ and $m\angle BCE$.
3. Find $m\widehat{BC}$. How does the measure of this arc relate to $m\angle BCE$?

Theorem 9-11: The measure of an angle formed by a chord and a tangent that intersect on the circle is half the measure of the intercepted arc.

$m\angle DBA=\frac{1}{2} m\widehat{AB}$

We now know that there are two types of angles that are half the measure of the intercepted arc; an inscribed angle and an angle formed by a chord and a tangent.

Example 1: Find:

a) $m\angle BAD$

b) $m \widehat{AEB}$

Solution: Use Theorem 9-11.

a) $m\angle BAD = \frac{1}{2} m \widehat{AB} = \frac{1}{2} \cdot 124^\circ=62^\circ$

b) $m\widehat{AEB} = 2 \cdot m\angle DAB= 2 \cdot 133^\circ=266^\circ$

Example 2: Find $a, \ b,$ and $c$.

Solution: $50^\circ + 45^\circ + m\angle a & = 180^\circ \qquad \text{straight angle}\\m\angle a & = 85^\circ$

$m \angle b & = \frac{1}{2} \cdot m \widehat{AC}\\m \widehat{AC} & = 2 \cdot m\angle EAC = 2 \cdot 45^\circ=90^\circ\\m \angle b & = \frac{1}{2} \cdot 90^\circ=45^\circ$

$85^\circ + 45^\circ + m\angle c & = 180^\circ \qquad \text{Triangle Sum Theorem}\\m\angle c & = 50^\circ$

From this example, we see that Theorem 9-8 is true for angles formed by a tangent and chord with the vertex on the circle. If two angles, with their vertices on the circle, intercept the same arc then the angles are congruent.

## Angles inside a Circle

An angle is inside a circle when the vertex anywhere inside the circle, but not on the center.

Investigation 9-7: Find the Measure of an Angle inside a Circle

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

1. Draw $\bigodot A$ with chord $\overline{BC}$ and $\overline{DE}$. Label the point of intersection $P$.
2. Draw central angles $\angle DAB$ and $\angle CAE$. Use colored pencils, if desired.
3. Find $m\angle DPB, \ m\angle DAB,$ and $m\angle CAE$. Find $m \widehat{DB}$ and $m \widehat{CE}$.
4. Find $\frac{m\widehat{DB} + m\widehat{CE}}{2}$.
5. What do you notice?

Theorem 9-12: The measure of the angle formed by two chords that intersect inside a circle is the average of the measure of the intercepted arcs.

$m\angle SVR & = \frac{1}{2} \left ( m\widehat{SR} + m\widehat{TQ} \right )= \frac{m\widehat{SR}+m\widehat{TQ}}{2}=m\angle TVQ\\m\angle SVT& = \frac{1}{2} \left (m\widehat{ST} + m\widehat{RQ} \right )=\frac{m\widehat{ST}+m\widehat{RQ}}{2} = m\angle RVQ$

Example 3: Find $x$.

a)

b)

c)

Solution: Use Theorem 9-12 to write an equation.

a) $x=\frac{129^\circ+71^\circ}{2}=\frac{200^\circ}{2}=100^\circ$

b) $40^\circ= \frac{52^\circ+x}{2}\!\\80^\circ=52^\circ+x\!\\28^\circ=x$

c) $x$ is supplementary to the angle that the average of the given intercepted arcs, $y$.

$y=\frac{19^\circ+107^\circ}{2}=\frac{126^\circ}{2}=63^\circ \qquad x+63^\circ=180^\circ; \ x=117^\circ$

## Angles outside a Circle

An angle is outside a circle if the vertex of the angle is outside the circle and the sides are tangents or secants. The possibilities are: an angle formed by two tangents, an angle formed by a tangent and a secant, and an angle formed by two secants.

Investigation 9-8: Find the Measure of an Angle outside a Circle

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

1. Draw three circles and label the centers $A, \ B,$ and $C$. In $\bigodot A$ draw two secant rays with the same endpoint. In $\bigodot B$, draw two tangent rays with the same endpoint. In $\bigodot C$, draw a tangent ray and a secant ray with the same endpoint. Label the points like the pictures below.
2. Draw in all the central angles. Using a protractor, measure the central angles and find the measures of each intercepted arc.
3. Find $m\angle EDF, \ m\angle MLN,$ and $m\angle RQS$.
4. Find $\frac{m\widehat{EF}-m\widehat{GH}}{2}, \ \frac{m\widehat{MPN}-m\widehat{MN}}{2},$ and $\frac{m\widehat{RS}-m\widehat{RT}}{2}$. What do you notice?

Theorem 9-13: The measure of an angle formed by two secants, two tangents, or a secant and a tangent from a point outside the circle is half the difference of the measures of the intercepted arcs.

$m\angle D & = \frac{m\widehat{EF}-m\widehat{GH}}{2}\\m\angle L & =\frac{m\widehat{MPN}-m\widehat{MN}}{2}\\m\angle Q & =\frac{m\widehat{RS}-m\widehat{RT}}{2}$

Example 4: Find the measure of $x$.

a)

b)

c)

Solution: For all of the above problems we can use Theorem 9-13.

a) $x=\frac{125^\circ-27^\circ}{2}=\frac{98^\circ}{2}=49^\circ$

b) $40^\circ$ is not the intercepted arc. The intercepted arc is $120^\circ, \ (360^\circ-200^\circ-40^\circ)$. $x=\frac{200^\circ-120^\circ}{2}=\frac{80^\circ}{2}=40^\circ$

c) Find the other intercepted arc, $360^\circ-265^\circ=95^\circ$ $x = \frac{265^\circ-95^\circ}{2}=\frac{170^\circ}{2}=85^\circ$

Know What? Revisited From Theorem 9-13, we know $x=\frac{182^\circ-178^\circ}{2}=\frac{4^\circ}{2}=2^\circ$.

## Review Questions

• Questions 1-3 use the definitions of tangent and secant lines.
• Questions 4-7 use the definition and theorems learned in this section.
• Questions 8-25 are similar to Examples 1-4.
• Questions 26 and 27 are similar to Example 4, but also a challenge.
• Questions 28 and 29 are fill-in-the-blank proofs of Theorems 9-12 and 9-13.
1. Draw two secants that intersect:
1. inside a circle.
2. on a circle.
3. outside a circle.
2. Can two tangent lines intersect inside a circle? Why or why not?
3. Draw a tangent and a secant that intersect:
1. on a circle.
2. outside a circle.

Fill in the blanks.

1. If the vertex of an angle is on the _______________ of a circle, then its measure is _______________ to the intercepted arc.
2. If the vertex of an angle is _______________ a circle, then its measure is the average of the __________________ arcs.
3. If the vertex of an angle is ________ a circle, then its measure is ______________ the intercepted arc.
4. If the vertex of an angle is ____________ a circle, then its measure is ___________ the difference of the intercepted arcs.

For questions 8-25, find the value of the missing variable(s).

1. $y \ne 60^\circ$

Challenge Solve for $x$.

1. Fill in the blanks of the proof for Theorem 9-12. Given: Intersecting chords $\overline{AC}$ and $\overline{BD}$. Prove: $m\angle a=\frac{1}{2} \left (m\widehat{DC}+m\widehat{AB}\right )$
Statement Reason
1. Intersecting chords $\overline{AC}$ and $\overline{BD}$.

2. Draw $\overline{BC}$

Construction
3. $m\angle DBC=\frac{1}{2} m\widehat{DC}\!\\m\angle ACB=\frac{1}{2} m\widehat{AB}$
4. $m\angle a=m\angle DBC+m\angle ACB$
5. $m\angle a=\frac{1}{2} m\widehat{DC}+\frac{1}{2} m\widehat{AB}$
1. Fill in the blanks of the proof for Theorem 9-13. Given: Secant rays $\overrightarrow{AB}$ and $\overrightarrow{AC}$ Prove: $m\angle a = \frac{1}{2} \left (m\widehat{BC}-m\widehat{DE} \right )$
Statement Reason
1. Intersecting secants $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

2. Draw $\overline{BE}$.

Construction
3. $m\angle BEC=\frac{1}{2} m\widehat{BC}\!\\m\angle DBE=\frac{1}{2} m\widehat{DE}$
5. $m\angle a+m\angle DBE=m\angle BEC$
6. Subtraction PoE
7. Substitution
8. $m\angle a=\frac{1}{2} \left (m\widehat{BC}-m\widehat{DE} \right )$

1. $m \angle OML = m \angle OPL = 90^\circ$ because a tangent line and a radius drawn to the point of tangency are perpendicular.
2. $165^\circ + m \angle OML + m \angle OPL + m \angle MLP = 360^\circ\!\\{\;} \qquad \quad \ \ 165^\circ + 90^\circ + 90^\circ + m \angle MLP = 360^\circ\!\\{\;} \qquad \qquad \qquad \qquad \qquad \quad \ \ \ m \angle MLP = 15^\circ$
3. $m\widehat{MNP} = 360^\circ - 165^\circ = 195^\circ$

8 , 9 , 10

Feb 22, 2012

Dec 11, 2014