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9.4: Inscribed Angles

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Find the measure of inscribed angles and the arcs they intercept.

Review Queue

We are going to use #14 from the homework in the previous section.

1. What is the measure of each angle in the triangle? How do you know?
2. What do you know about the three arcs?
3. What is the measure of each arc?

Know What? The closest you can get to the White House are the walking trails on the far right. You want to get as close as you can (on the trail) to the fence to take a picture (you were not allowed to walk on the grass). Where else can you take a picture from to get the same frame of the White House? Your line of sight in the camera is marked in the picture as the grey lines.

Inscribed Angles

In addition to central angles, we will now learn about inscribed angles in circles.

Inscribed Angle: An angle with its vertex on the circle and sides are chords.

Intercepted Arc: The arc that is inside the inscribed angle and endpoints are on the angle.

The vertex of an inscribed angle can be anywhere on the circle as long as its sides intersect the circle to form an intercepted arc.

Investigation 9-4: Measuring an Inscribed Angle

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

1. Draw three circles with three different inscribed angles. Try to make all the angles different sizes.
2. Using your ruler, draw in the corresponding central angle for each angle and label each set of endpoints.
3. Using your protractor measure the six angles and determine if there is a relationship between the central angle, the inscribed angle, and the intercepted arc.

mLAM= mLMˆ= mLKM= mNBP= mNPˆ= mNOP= mQCR= mQRˆ=mQSR= \begin{align*}& m\angle LAM = \ \underline{\;\;\;\;\;\;\;\;\;\;} && m\angle NBP= \ \underline{\;\;\;\;\;\;\;\;\;\;} && m\angle QCR = \ \underline{\;\;\;\;\;\;\;\;\;\;}\\ & m\widehat{LM} = \ \underline{\;\;\;\;\;\;\;\;\;\;} && m\widehat{NP} = \ \underline{\;\;\;\;\;\;\;\;\;\;} && m\widehat{QR}= \underline{\;\;\;\;\;\;\;\;\;\;}\\ & m\angle LKM = \ \underline{\;\;\;\;\;\;\;\;\;\;} && m\angle NOP = \ \underline{\;\;\;\;\;\;\;\;\;\;} && m\angle QSR = \ \underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}

Inscribed Angle Theorem: The measure of an inscribed angle is half the measure of its intercepted arc.

mADCmACˆ=12mACˆ=2mADC\begin{align*}m\angle ADC& =\frac{1}{2} m \widehat{AC}\\ m\widehat{AC} & = 2 m\angle ADC\end{align*}

Example 1: Find mDCˆ\begin{align*}m \widehat{DC}\end{align*} and mADB\begin{align*}m\angle ADB\end{align*}.

Solution: From the Inscribed Angle Theorem:

mDCˆmADB=245=90=1276=38\begin{align*}m \widehat{DC} & = 2 \cdot 45^\circ=90^\circ \\ m\angle ADB & = \frac{1}{2}\cdot 76^\circ=38^\circ\end{align*}

Example 2: Find mADB\begin{align*}m\angle ADB\end{align*} and mACB\begin{align*}m\angle ACB\end{align*}.

Solution: The intercepted arc for both angles is ABˆ\begin{align*}\widehat{AB}\end{align*}. Therefore,

mADBmACB=12124=62=12124=62\begin{align*}m\angle ADB & = \frac{1}{2} \cdot 124^\circ=62^\circ\\ m\angle ACB & = \frac{1}{2} \cdot 124^\circ=62^\circ\end{align*}

This example leads us to our next theorem.

Theorem 9-8: Inscribed angles that intercept the same arc are congruent.

ADB\begin{align*}\angle ADB\end{align*} and ACB\begin{align*}\angle ACB\end{align*} intercept ABˆ\begin{align*}\widehat{AB}\end{align*}, so mADB=mACB\begin{align*}m\angle ADB = m\angle ACB\end{align*}.

DAC\begin{align*}\angle DAC\end{align*} and DBC\begin{align*}\angle DBC\end{align*} intercept DCˆ\begin{align*}\widehat{DC}\end{align*}, so mDAC=mDBC\begin{align*}m\angle DAC = m\angle DBC\end{align*}.

Example 3: Find mDAB\begin{align*}m\angle DAB\end{align*} in \begin{align*}\bigodot C\end{align*}.

Solution: \begin{align*}C\end{align*} is the center, so \begin{align*}\overline{DB}\end{align*} is a diameter. \begin{align*}\angle DAB\end{align*} endpoints are on the diameter, so the central angle is \begin{align*}180^\circ\end{align*}.

\begin{align*}m\angle DAB & = \frac{1}{2} \cdot 180^\circ=90^\circ.\end{align*}

Theorem 9-9: An angle intercepts a semicircle if an only if it is a right angle.

\begin{align*}\angle DAB\end{align*} intercepts a semicircle, so \begin{align*}m\angle DAB = 90^\circ\end{align*}.

\begin{align*}\angle DAB\end{align*} is a right angle, so \begin{align*}\widehat{DB}\end{align*} is a semicircle.

Anytime a right angle is inscribed in a circle, the endpoints of the angle are the endpoints of a diameter and the diameter is the hypotenuse.

Example 4: Find \begin{align*}m\angle PMN, \ m\widehat{PN}, \ m\angle MNP,\end{align*} and \begin{align*}m\angle LNP\end{align*}.

Solution:

\begin{align*}m\angle PMN & = m\angle PLN=68^\circ \quad \ \ \text{by Theorem} \ 9-8.\\ m\widehat{PN} & = 2 \cdot 68^\circ=136^\circ \quad \ \ \ \text{from the Inscribed Angle Theorem.}\\ m\angle MNP & =90^\circ \qquad \qquad \qquad \ \text{by Theorem} \ 9-9.\\ m\angle LNP & = \frac{1}{2} \cdot 92^\circ=46^\circ \qquad \ \text{from the Inscribed Angle Theorem.}\end{align*}

Inscribed Polygon: A polygon where every vertex is on a circle.

Tools Needed: pencil, paper, compass, ruler, colored pencils, scissors

1. Draw a circle. Mark the center point \begin{align*}A\end{align*}.
2. Place four points on the circle. Connect them to form a quadrilateral. Color in the 4 angles.
3. Cut out the quadrilateral. Then cut the diagonal \begin{align*}\overline{CE}\end{align*}, making two triangles.
4. Line up \begin{align*}\angle B\end{align*} and \begin{align*}\angle D\end{align*} so that they are next to each other. What do you notice?

By cutting the quadrilateral in half, we are able to show that \begin{align*}\angle B\end{align*} and \begin{align*}\angle D\end{align*} form a linear pair when they are placed next to each other, making \begin{align*}\angle B\end{align*} and \begin{align*}\angle D\end{align*} supplementary.

Theorem 9-10: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary.

If \begin{align*}ABCD\end{align*} is inscribed in \begin{align*}\bigodot E\end{align*}, then \begin{align*}m\angle A+m\angle C=180^\circ\end{align*} and \begin{align*}m\angle B+m\angle D=180^\circ\end{align*}.

If \begin{align*}m\angle A+m\angle C=180^\circ\end{align*} and \begin{align*}m\angle B+m\angle D=180^\circ\end{align*}, then \begin{align*}ABCD\end{align*} is inscribed in \begin{align*}\bigodot E\end{align*}.

Example 5: Find the value of the missing variables.

a)

b)

Solution:

a) \begin{align*}x+80^\circ=180^\circ \qquad \qquad y+71^\circ=180^\circ\!\\ x =100^\circ \qquad \qquad \qquad \qquad \qquad \ \ y=109^\circ\end{align*}

b) \begin{align*}z+93^\circ =180^\circ \qquad \qquad x=\frac{1}{2} (58^\circ+106^\circ) \qquad \qquad y+82^\circ=180^\circ\!\\ z=87^\circ \qquad \qquad \qquad \qquad \ x=82^\circ \qquad \qquad \qquad \qquad \quad \qquad \ y=98^\circ\end{align*}

Example 6: Algebra Connection Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the picture below.

Solution:

\begin{align*}(7x+1)^\circ+105^\circ& =180^\circ && (4y+14)^\circ+(7y+1)^\circ=180^\circ\\ 7x+106^\circ&=180^\circ && \qquad \qquad \quad \ 11y+15^\circ=180^\circ\\ 7x&=84^\circ && \qquad \qquad \qquad \qquad 11y=165^\circ\\ x&=12^\circ && \qquad \qquad \qquad \qquad \quad y=15^\circ\end{align*}

Know What? Revisited You can take the picture from anywhere on the semicircular walking path, the frame will be the same.

Review Questions

• Questions 1-8 use the vocabulary and theorems learned in this section.
• Questions 9-27 are similar to Examples 1-5.
• Questions 28-33 are similar to Example 6.
• Question 34 is a proof of the Inscribed Angle Theorem.

Fill in the blanks.

1. A\begin{align*}(n)\end{align*} _______________ polygon has all its vertices on a circle.
2. An inscribed angle is ____________ the measure of the intercepted arc.
3. A central angle is ________________ the measure of the intercepted arc.
4. An angle inscribed in a ________________ is \begin{align*}90^\circ\end{align*}.
5. Two inscribed angles that intercept the same arc are _______________.
6. The _____________ angles of an inscribed quadrilateral are ________________.
7. The sides of an inscribed angle are ___________________.
8. Draw inscribed angle \begin{align*}\angle JKL\end{align*} in \begin{align*}\bigodot M\end{align*}. Then draw central angle \begin{align*}\angle JML\end{align*}. How do the two angles relate?

Quadrilateral \begin{align*}ABCD\end{align*} is inscribed in \begin{align*}\bigodot E\end{align*}. Find:

1. \begin{align*}m\angle DBC\end{align*}
2. \begin{align*}m \widehat{BC}\end{align*}
3. \begin{align*}m \widehat{AB}\end{align*}
4. \begin{align*}m\angle ACD\end{align*}
5. \begin{align*}m\angle ADC\end{align*}
6. \begin{align*}m\angle ACB\end{align*}

Quadrilateral \begin{align*}ABCD\end{align*} is inscribed in \begin{align*}\bigodot E\end{align*}. Find:

1. \begin{align*}m\angle A\end{align*}
2. \begin{align*}m\angle B\end{align*}
3. \begin{align*}m\angle C\end{align*}
4. \begin{align*}m\angle D\end{align*}

Find the value of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*} in \begin{align*}\bigodot A\end{align*}.

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

1. Fill in the blanks of the Inscribed Angle Theorem proof. Given: Inscribed \begin{align*}\angle ABC\end{align*} and diameter \begin{align*}\overline{BD}\end{align*} Prove: \begin{align*}m\angle ABC = \frac{1}{2} m \widehat{AC}\end{align*}
Statement Reason

1. Inscribed \begin{align*}\angle ABC\end{align*} and diameter \begin{align*}\overline{BD}\end{align*}

\begin{align*}m\angle ABE = x^\circ\end{align*} and \begin{align*}m\angle CBE = y^\circ\end{align*}

2. \begin{align*}x^\circ + y^\circ = m\angle ABC\end{align*}
4. Definition of an isosceles triangle
5. \begin{align*}m\angle EAB = x^\circ\end{align*} and \begin{align*}m\angle ECB = y^\circ\end{align*}
6. \begin{align*}m\angle AED = 2x^\circ\end{align*} and \begin{align*}m\angle CED = 2y^\circ\end{align*}
7. \begin{align*}m\widehat{AD}= 2x^\circ\end{align*} and \begin{align*}m \widehat{DC} = 2y^\circ\end{align*}
9. \begin{align*}m\widehat{AC} = 2x^\circ + 2y^\circ\end{align*}
10. Distributive PoE
11. \begin{align*}m\widehat{AC} = 2m\angle ABC\end{align*}
12. \begin{align*}m\angle ABC=\frac{1}{2} m \widehat{AC}\end{align*}

1. \begin{align*}60^\circ\end{align*}, it is an equilateral triangle.
2. They are congruent because the chords are congruent.
3. \begin{align*}\frac{360^\circ}{3} = 120^\circ\end{align*}

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