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CHAPTER7.5Y
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à 7.5èInscribed Angle Measure
äèPlease answer ê followïg questions about ïscribed
angle measure.
â
èèèèèèè The measure ç an ïscribed angle is oneè
èèèèèèè half ê measure ç ê arc it ïtercepts.
éS1
Defïition 7.5.1èINSCRIBED ANGLE:èAn ïscribed angle is an angle whose
vertex is on a circle å whose sides ïtersect ê circle ï two oêr
poïts.
èèèèèèèèèèèèèèèèèèèèè In this figure, ╬BAC is an
èèèèèèèèèèèèèèèèèèèèè ïscribed angle for circle
èèèèèèèèèèèèèèèèèèèèè P.èArc BC is ê arc ï-
èèèèèèèèèèèèèèèèèèèèè tercepted by ╬BAC.è╬BPC
@fig7501.BMP,85,110,147,74èèèèèèèè is a central angle for
circle P.èThe measure ç ê central angle, ╬BPC, is also ê measure ç
arc BC.èThe followïg êorem gives ê relationship between ê mea-
sure ç ê ïscribed angle, ╬BAC, å ê measure ç ê ïtercepted
arc.
Theorem 7.5.1èThe measure ç an ïscribed angle is one half ê measure
ç its ïtercepted arc.è
Theorem 7.5.2èIf two ïscribed angles ïtercept ê same arc, ên ê
angles are equal.
Proç: For a proç please see Problem 1.
Theorem 7.5.3èIf an angle is ïscribed ï a semicircle, ên it is a
right angle.
Proç: For a proç please see Problem 2.
Theorem 7.5.4èIf a quadrilateral is ïscribed ï a circle, ên its
opposite angles are supplementary.
Proç: For a proç please see Problem 3.
1èèèèèèèèèèèè If ╬ACB å ╬AEB are ïscribed angles
èèèèèèèèèèèèèèèèèthat ïtercept ê same arc, arc AB,
èèèèèèèèèèèèèèèèècan you show that ╬ACB ╧ ╬AEB?
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè A) Yesèèèè B) No
@fig7502.BMP,35,40,147,74èè
ü Show ╬ACB ╧ ╬AEB
Proç: StatementèèèèèèèèèèèReason
èèè 1. m╬ACB = 1/2 m arc ABèèèè1. Theorem 7.5.1
èèè 2. m╬AEB = 1/2 m arc ABèèèè2. Theorem 7.5.1
èèè 3. m╬ACB = m╬AEBèèèèèèè 3. Transitive axiom for equality
èèè 4. ╬ACB ╧ ╬AEBèèèèèèèè 4. Defïition ç congruent
Ç A
2èèèèèèèèèèèèèèIf ╖┤ is a diameter å ╬BAC is
èèèèèèèèèèèèèèèèèè an ïscribed angle, can you show
èèèèèèèèèèèèèèèèèè that m╬BAC = 90°?
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèè A) Yesèèèè B) No
@fig7503.BMP,35,40,147,74èè
ü Show m╬BAC = 90°
Proç: Statementèèèèèèèèèèèè Reason
èèè 1. ╬BAC is ïscribedèèèèèèè1. Given
èèè 2. m╬BAC = 1/2 m arc BECèèèèè2. Theorem 7.5.1
èèè 3. m arc BEC = 180°èèèèèèè 3. Semicircle is 180°
èèè 4. m╬BAC = 1/2·180°èèèèèèè 4. Substitution
èèè 5. m╬BAC = 90°èèèèèèèèèè5. Multiplication
Ç A
3èèèèèèèèèèèèè If quadrilateral ABCE is ïscribed
èèèèèèèèèèèèèèèèèèï circle P, can you show that
èèèèèèèèèèèèèèèèèèm╬ABC + m╬AEC = 180°?
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèè A) Yesèèèè B) No
@fig7504.BMP,35,40,147,74èè
ü Show m╬ABC + m╬AEC = 180°
Proç: Statementèèèèèèèèèèèè Reason
èèè 1. ╬ABC, ╬AEC are ïscribed ╬sèè1. Given
èèè 2. m╬ABC = 1/2 m arc CEAèèèèè2. Theorem 7.5.1
èèè 3. m╬AEC = 1/2èm arc ABCèèèè 3. Theorem 7.5.1
èèè 4. m╬ABC + m╬AEC =èèèèèèèè4. Addition axiom for equations
èèèèèè1/2(m arc CEA + m arc ABC)èè
èèè 5. m╬ABC + m╬AEC = 1/2(360°)èèè5. Circle has 360°
èèè 6. m╬ABC + m╬AEC = 180°èèèèè 6. Multiplication
Ç A
4èèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèName ê arc ïtercepted byè
èèèèèèèèèèèèèèèèèê ïscribed angle, ╬CBA.
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèè A) arc EABèè B) arc ABCèè C) arc CEA
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèèThe ïtercepted arc is arc CEA.
Ç C
5èèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè If m arc ABC is 170°,èè
èèèèèèèèèèèèèèèèèèèè fïd m╬AEC.
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèA) 85°èè B) 95°èè C) 105°
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèm╬AEC = 1/2 m arc ABC = 1/2·170° = 85°
Ç A
6èèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèèè If m╬BCE = 60°,èè
èèèèèèèèèèèèèèèèèèèèèè fïd m╬BAE.
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèA) 360°èè B) 120°èè C) 90°
@fig7504.BMP,35,40,147,74èè
ü
èè Opposite angles ç an ïscribed quadrilateral are supplementary.
èèèèèèèèèèèèèè 180° - 60° = 120°èèèèè
Ç B
7èèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèè Name ê ïscribed angleèè
èèèèèèèèèèèèèèèèèèè that ïtercepts arc BAE.
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèA) ╬BCEèè B) ╬ABCèè C) ╬BAE
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèèèè╬BCE ïtercepts arc BAE.èèèèèèèè
Ç A
8èèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèName an ïscribed angleèè
èèèèèèèèèèèèèèèèèèèèthat is congruent ë ╬CAE.
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèè A) ╬BACèè B) ╬CBEèè C) ╬ABE
@fig7504.BMP,35,40,147,74èè
ü
èèèèèTwo angles that ïtercept ê same arc are congruent.
èèèè╬CBE ïtercepts arc CE which is also ïtercepted by ╬CAE.èèèèèèèèèèèèèèè
Ç B
9èèèèèèèèè
èèèèèèèèèèèèèèèèèèè If m╬AEB = 35°, fïd m╬BCA.èè
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèèA) 45°èè B) 35°èè C) 55°
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèSïce ╬AEB å ╬BCA ïtercept ê sameè
èèèèèèèèèarc, arc AB,êir measures are equal.
èè
èèèèèèèèèèèèèèè m╬BCA = 35°èèèèèèèèèèèèèèèè
Ç B
10èèèèèèèèè
èèèèèèèèèèèèèèèèèèè If m arc BC = 50°, fïd m╬BAC.èè
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè A) 25°èè B) 30°èè C) 45°
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèèm╬BAC = 1/2 m arc BC = 1/2·50° = 25°èèèèèèèèèèèèèèèèèèèèèèè
Ç A
11èèèèèèèèè
èèèèèèèèèèèèèèèèèèè If m╬ECA = 20°, fïd m arc AE.èè
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè A) 60°èè B) 40°èè C) 45°
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèèèm arc AE = 2m╬ECA = 2·20° = 40°èèèèèèèèèèèèèèèèèèèèèèè
Ç B
12èèèèèèèèè
èèèèèèèèèèèèèèèèèèèè If m╬BAC = 35°, fïd m╬BEC.èè
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèèè A) 45°èè B) 70°èè C) 35°
@fig7504.BMP,35,40,147,74èè
ü
èèèèèèèèèè Sïce ╬BAC å ╬BEC both ïtercept
èèèèèèèèèè arc BC, êy have equal measures.
èèèèèèèèèèèèèèèèm╬BEC = 35°èè
Ç C
13èèèèèèèèè
èèèèèèèèèèèèèèèèèèIf ╖┤ is a diameter, fïd m╬BAC.èè
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèè A) 90°èè B) 180°èè C) 60°
@fig7503.BMP,35,40,147,74èè
ü
èèèèèAn angle ïscribed ï a semicircle is a right angle.èèèèèèè
Ç A
14èèèèèèèèè
èèèèèèèèèèèèèèèèèèè If m╬BPC = 120°, fïd m╬BAC.èè
èèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèèè A) 60°èè B) 120°èè C) 90°
@fig7501.BMP,35,40,147,74èè
ü
èèèèèèèm╬BAC = 1/2 m arc BC = 1/2 m╬BAC = 1/2·120° = 60°èèèèèèè
Ç A