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To help you understand it, consider using the figure below to illustrate our lesson. Some have already been completed for you. So, keep doing the others. You can select your answer from the box below. Major Arc: BDA, Minor Arc: AE B Semicircle: BAE, Central Angles: LACE. *Suppose m LAFE = 95°, then m AE = 95° %3D %3D * If m AB = 105°, then m LACB = *Suppose mLAFE = 95°, find m ADB = Given: AE= 95°, EDB = 180° (Semicircle) Solution: ZAFE = AÈ %3! A %3D MADB = AE + EDB = 95° + 180° = (2759 D *If MLACB =105°, then m AKE =

To help you understand it, consider using the figure below to illustrate our lesson. Some have already been completed for you. So, keep doing the others. You can select your answer from the box below. Major Arc: BDA, Minor Arc: AE B Semicircle: BAE, Central Angles: LACE. *Suppose m LAFE = 95°, then m AE = 95° %3D %3D * If m AB = 105°, then m LACB = *Suppose mLAFE = 95°, find m ADB = Given: AE= 95°, EDB = 180° (Semicircle) Solution: ZAFE = AÈ %3! A %3D MADB = AE + EDB = 95° + 180° = (2759 D *If MLACB =105°, then m AKE =

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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v To help you understand it, consider using the figure below to illustrate our lesson. Some have already been completed for you. So, keep doing the others. You can select your answer from the box below. Major Arc: BDA, Minor Arc: AE B Semicircle: BAE, Central Angles: 4ACE. *Suppose m LAFE = 95°, then m AE = 95° %3D If m AB = 105°, then m LACB = 1. *Suppose mLAFE = 95°, find m ADB = Given: AE= 95°, EDB = 180° (Semicircle) Solution: ZAFE =AÈ %3D A %3D MADB = AE + EDB = 95° + 180 ° = (2759 D *If MLACB =105°, then m AKE = K *If MZBCA = 84° and mLEFA = 96°, find m BE Solution: m BE = mzBCA+ MZEFA m BE = m BA + m EA = 84° + 96° =(18:0 E BEA ВАК ВЕК ВКЕ BDE ВКА BA KE *If MZKFE = 37° and mLEFA= 58°, find m KA ZBCA ZKFE ZKBE 105° 285° 95° %3D 180° BE
Transcribed Image Text:v To help you understand it, consider using the figure below to illustrate our lesson. Some have already been completed for you. So, keep doing the others. You can select your answer from the box below. Major Arc: BDA, Minor Arc: AE B Semicircle: BAE, Central Angles: 4ACE. *Suppose m LAFE = 95°, then m AE = 95° %3D If m AB = 105°, then m LACB = 1. *Suppose mLAFE = 95°, find m ADB = Given: AE= 95°, EDB = 180° (Semicircle) Solution: ZAFE =AÈ %3D A %3D MADB = AE + EDB = 95° + 180 ° = (2759 D *If MLACB =105°, then m AKE = K *If MZBCA = 84° and mLEFA = 96°, find m BE Solution: m BE = mzBCA+ MZEFA m BE = m BA + m EA = 84° + 96° =(18:0 E BEA ВАК ВЕК ВКЕ BDE ВКА BA KE *If MZKFE = 37° and mLEFA= 58°, find m KA ZBCA ZKFE ZKBE 105° 285° 95° %3D 180° BE
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