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°.
In the above diagram,
m∠arc PRQ = 180°
Example 1 :
From the diagram shown above, find the following arc measures.
(i) m∠arc BC
(ii) m∠arc ABC
Solution :
(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°
= 237°
Example 2 :
From the diagram shown above, find the following measures.
(i) m∠arc CD
(ii) m∠AOC
(iii) m∠arc BD
(iv) m∠arc ABC
(v) m∠arc CBD
Solution :
(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°
= 235°
(v) m∠arc CBD :
m∠arc CBD = m∠arc CAB + m∠arc BD
= 180° + 125°
= 305°
Example 3 :
Find the value of x in the diagram shown below.
From the diagram shown above, find the m∠arc QTR.
Solution :
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°
= 306°
Example 4 :
From the diagram shown above, find the following measures.
m∠BOD, m∠BOE and m∠BOC
Solution :
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°
= 146°
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 :
From the diagram shown above, find the following measures.
m∠KOL and m∠arc MNK
Solution :
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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