homework 2 central angles & arc measures answers

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Homework 2 Central Angles Arc Measures

Homework 2 Central Angles Arc Measures - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Assignment, Arc length and sector area, Geometry 10 2 angles and arcs, 11 arcs and central angles, Geometry unit 10 notes circles, 11 arcs and central angles, Homework section 9 1.

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1. Assignment

2. arc length and sector area, 3. geometry 10-2 angles and arcs, 4. 11-arcs and central angles, 5. geometry unit 10 notes circles, 6. 11-arcs and central angles, 7. homework section 9-1.

Algebra and Pre-Algebra

Measures of arcs and central angles

We can use a few more theorems to find the measures of arcs and central angles of circles.  Let’s begin by stating a few theorems:

THEOREM:   The measure of a central angle is equal to the measure of the arc it intersects.

THEOREM:   The measure of a major arc (an arc greater than a semicircle) is equal to \(360^\circ \) minus the measure of the corresponding minor arc.

THEOREM:   Vertical angles are equal.

EXAMPLE:   Find the measure of the arc \(\widehat {IKH}\)

SOLUTION: \(\widehat {IKH}\) is a major arc, so, by a theorem above, its measure is \(360^\circ  - m\widehat {IH}\). Then we must find the measure of \(\widehat {IH}\).

Again, by a theorem above, we know that the measure of the central angle corresponding to \(\widehat {JK}\) must be \(70^\circ \). Now we can conclude, by the fact that straight angles measure \(180^\circ \), that the central angle corresponding to \(\widehat {IH}\) equals \(180 - 70 - 70 = 40\). That is, \(m\widehat {IH} = 40^\circ \).

So we can conclude that \(m\widehat {IKH} = 360^\circ  - 40^\circ  = 320^\circ \).

EXAMPLE:   Find \(m\angle JKI\)

SOLUTION:   Since \(\widehat {FKG} = 50^\circ \), we can conclude, by a theorem above, that \(\angle FKG = 50^\circ \). Then, by the vertical angle theorem, we know that \(\angle JKL = \angle FKG\). That is, \(\angle JKL = 50^\circ \).

Below you can download some free math worksheets and practice.

Find the measure of the arc or central angle indicated. Assume that lines which appear to be diameters are actual diameters.

This free worksheet contains 10 assignments each with 24 questions with answers. Example of one question:

Watch bellow how to solve this example:

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CENTRAL ANGLES AND ARC MEASURES

1. A central angle is an angle with its vertex at the center of the circle and its two sides are radii. 

2. For example : m ∠POQ is a central angle in circle P shown below. 

3. The sum of all central angle is 360 °.

4. The measure of the arc formed by the endpoints of a central angle is equal to the degree of the central angle.

In the above diagram, 

m ∠arc PQ = 85 °

m ∠arc PRQ = 360 ° - 85 ° = 275 °

5. The measure of the arc formed by the endpoints of the diameter is equal to 180 ° .

m∠arc PRQ = 180 °

Example 1 : 

From the diagram shown above, find the following arc measures. 

(i)  m ∠arc BC

(ii) m∠arc ABC

(i)  m ∠arc BC :

AB is the diameter of the above circle. 

m∠arc AB = 180 °

m∠arc BC +  m∠arc CA = 180 °

m∠arc BC + 123 °  = 180 °

m∠arc BC  = 57 °

(ii) m∠arc ABC :

m∠arc ABC = m ∠arc AB + m ∠arc BC

= 180 °  + 57 °

Example 2 :

From the diagram shown above, find the following measures. 

(i)  m ∠arc CD

(iii) m∠arc BD

(iv) m∠arc ABC

(v) m∠arc CBD

(i)  m ∠arc CD :

m∠AOB and m ∠COD are vertical angles. 

m ∠COD = m ∠AOB

m ∠arc CD = m ∠arc AB

m∠arc CD = 55 °

(ii) m∠AOC :

BC is the diameter of the above circle. 

m∠arc BAC = 180 °

m∠arc BA +  m∠arc AC = 180 °.

55 °  +  m∠arc AC = 180 °.

m∠arc AC = 125 °.

m∠AOC = 125 °.

(iii) m∠arc BD : 

m∠BOD and m ∠AOC are vertical angles. 

m ∠BOD = m ∠AOC

m ∠BOD = 125 °

m∠arc BD = 125°

(iv) m∠arc ABC : 

m∠arc ABC =  m∠arc ABD +  m∠arc DC

= 180 °  + 55 °

(v) m∠arc CBD : 

m∠arc CBD =  m∠arc CAB +  m∠arc BD

= 180 °  + 125 °

Example 3 :

Find the value of x in the diagram shown below. 

From the diagram shown above, find the  m ∠arc QTR.

Find m ∠arc QP :

PS is the diameter of the above circle.

m ∠arc PTS = 180 °

m∠arc PT +  m∠arc TS  = 180°

135 ° +  m∠arc TS  = 180°

m∠arc TS = 45°

Find m ∠arc QTR :

m∠QTR = m ∠arc QT + m ∠arc TS + m ∠arc SR

= 180 ° + 45 ° + 81 °

Example 4 :

m ∠BOD,   m ∠BOE and  m ∠BOC

Find  m ∠BOD :

In the circle above,

m ∠arc AB +  m ∠arc BCD +  m ∠arc DE +  m ∠arc EA = 360 °

60 °  +  m ∠arc BCD + 86 °  + 154 °  = 360 °

m ∠arc BCD + 300 °  = 360 °

m ∠arc BCD  = 60 °

m ∠BOD  = 60 °

Find  m ∠BOE :

m ∠BOE = m ∠arc BCD + m∠arc DE

= 60 ° + 86 °

Find m ∠BOC :

In the above diagram,  m∠BOC =  m ∠COD.

m∠BOC + m∠COD =  m∠BOD

m∠BOC + m∠BOC = m∠BOD

2m∠BOC = 60 °

m∠BOC = 30 °

Example 5 :

m ∠ KOL and  m∠arc MNK

In the diagram above,  m∠JON and  ∠KOM are vertical angles.

m∠KOM  = m ∠KOM

m∠KOM = 126 °

m∠KOL + m ∠LOM  = 126 °

In the above diagram,  m∠KOL =  m ∠LOM.

m∠KOL + m∠KOL = 126°

2m∠KOL = 126°

m ∠ KOL = 63°

Find m ∠arc MNK :

m∠arc MNK = 360 ° - m ∠arc KLM

m∠arc MNK = 360° - m∠KOM

m∠arc MNK = 360° - 126 °

m∠arc MNK = 234 °

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What is the measure of ∠PTQ?

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