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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 mOML\begin{align*}m\angle OML\end{align*} and mOPL\begin{align*}m\angle OPL\end{align*}? How do you know?
2. Find mMLP\begin{align*}m\angle MLP\end{align*}.
3. Find mMNPˆ\begin{align*}m\widehat{MNP}\end{align*}.

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\begin{align*}178^\circ\end{align*}, what is the angle at which the sun’s rays hit the earth (x)\begin{align*}(x^\circ)\end{align*}?

## 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 A\begin{align*}\bigodot A\end{align*} with chord BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and tangent line ED\begin{align*}\overleftrightarrow{ED}\end{align*} with point of tangency C\begin{align*}C\end{align*}.
2. Draw in central angle CAB\begin{align*}\angle CAB\end{align*}. Find mCAB\begin{align*}m\angle CAB\end{align*} and mBCE\begin{align*}m\angle BCE\end{align*}.
3. Find mBCˆ\begin{align*}m\widehat{BC}\end{align*}. How does the measure of this arc relate to mBCE\begin{align*}m\angle BCE\end{align*}?

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.

mDBA=12mABˆ\begin{align*}m\angle DBA=\frac{1}{2} m\widehat{AB}\end{align*}

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) mBAD\begin{align*}m\angle BAD\end{align*}

b) mAEBˆ\begin{align*}m \widehat{AEB}\end{align*}

Solution: Use Theorem 9-11.

a) mBAD=12mABˆ=12124=62\begin{align*}m\angle BAD = \frac{1}{2} m \widehat{AB} = \frac{1}{2} \cdot 124^\circ=62^\circ\end{align*}

b) mAEBˆ=2mDAB=2133=266\begin{align*}m\widehat{AEB} = 2 \cdot m\angle DAB= 2 \cdot 133^\circ=266^\circ\end{align*}

Example 2: Find a, b,\begin{align*}a, \ b,\end{align*} and c\begin{align*}c\end{align*}.

Solution: 50+45+mama=180straight angle=85\begin{align*}50^\circ + 45^\circ + m\angle a & = 180^\circ \qquad \text{straight angle}\\ m\angle a & = 85^\circ\end{align*}

mbmACˆmb=12mACˆ=2mEAC=245=90=1290=45\begin{align*}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\end{align*}

85+45+mcmc=180Triangle Sum Theorem=50\begin{align*}85^\circ + 45^\circ + m\angle c & = 180^\circ \qquad \text{Triangle Sum Theorem}\\ m\angle c & = 50^\circ\end{align*}

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 A\begin{align*}\bigodot A\end{align*} with chord BC¯¯¯¯¯¯¯¯\begin{align*}\overline{BC}\end{align*} and DE¯¯¯¯¯¯¯¯\begin{align*}\overline{DE}\end{align*}. Label the point of intersection P\begin{align*}P\end{align*}.
2. Draw central angles DAB\begin{align*}\angle DAB\end{align*} and CAE\begin{align*}\angle CAE\end{align*}. Use colored pencils, if desired.
3. Find mDPB, mDAB,\begin{align*}m\angle DPB, \ m\angle DAB,\end{align*} and mCAE\begin{align*}m\angle CAE\end{align*}. Find mDBˆ\begin{align*}m \widehat{DB}\end{align*} and mCEˆ\begin{align*}m \widehat{CE}\end{align*}.
4. Find mDBˆ+mCEˆ2\begin{align*}\frac{m\widehat{DB} + m\widehat{CE}}{2}\end{align*}.
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.

mSVRmSVT=12(mSRˆ+mTQˆ)=mSRˆ+mTQˆ2=mTVQ=12(mSTˆ+mRQˆ)=mSTˆ+mRQˆ2=mRVQ\begin{align*}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\end{align*}

Example 3: Find \begin{align*}x\end{align*}.

a)

b)

c)

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

a) \begin{align*}x=\frac{129^\circ+71^\circ}{2}=\frac{200^\circ}{2}=100^\circ\end{align*}

b) \begin{align*}40^\circ= \frac{52^\circ+x}{2}\!\\ 80^\circ=52^\circ+x\!\\ 28^\circ=x\end{align*}

c) \begin{align*}x\end{align*} is supplementary to the angle that the average of the given intercepted arcs, \begin{align*}y\end{align*}.

\begin{align*}y=\frac{19^\circ+107^\circ}{2}=\frac{126^\circ}{2}=63^\circ \qquad x+63^\circ=180^\circ; \ x=117^\circ\end{align*}

## 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 \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. In \begin{align*}\bigodot B\end{align*}, draw two tangent rays with the same endpoint. In \begin{align*}\bigodot C\end{align*}, 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 \begin{align*}m\angle EDF, \ m\angle MLN,\end{align*} and \begin{align*}m\angle RQS\end{align*}.
4. 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?

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.

\begin{align*}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}\end{align*}

Example 4: Find the measure of \begin{align*}x\end{align*}.

a)

b)

c)

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

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

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

c) 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*}

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

## 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. \begin{align*}y \ne 60^\circ\end{align*}

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

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

2. Draw \begin{align*}\overline{BC}\end{align*}

Construction
3. \begin{align*}m\angle DBC=\frac{1}{2} m\widehat{DC}\!\\ m\angle ACB=\frac{1}{2} m\widehat{AB}\end{align*}
4. \begin{align*}m\angle a=m\angle DBC+m\angle ACB\end{align*}
5. \begin{align*}m\angle a=\frac{1}{2} m\widehat{DC}+\frac{1}{2} m\widehat{AB}\end{align*}
1. Fill in the blanks of the proof for Theorem 9-13. 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*}
Statement Reason
1. Intersecting secants \begin{align*}\overrightarrow{AB}\end{align*} and \begin{align*}\overrightarrow{AC}\end{align*}.

2. Draw \begin{align*}\overline{BE}\end{align*}.

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

1. \begin{align*}m \angle OML = m \angle OPL = 90^\circ\end{align*} because a tangent line and a radius drawn to the point of tangency are perpendicular.
2. \begin{align*}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\end{align*}
3. \begin{align*}m\widehat{MNP} = 360^\circ - 165^\circ = 195^\circ\end{align*}

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