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Prove: m/COD B DA 72° Complete the proof. It is given that m/AOB = 42° and m/EOF = 66°. By the m/BOC = 66°. By the m/AOC + m/COD = 180°. After application of the m/AOC = 108°, and by the ZEOF ZBOC. Therefore, m/COD = 72°

Prove: m/COD B DA 72° Complete the proof. It is given that m/AOB = 42° and m/EOF = 66°. By the m/BOC = 66°. By the m/AOC + m/COD = 180°. After application of the m/AOC = 108°, and by the ZEOF ZBOC. Therefore, m/COD = 72°

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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Prove: m/COD = 72° KA Complete the proof. It is given that m/AOB m/BOC = 66°. By the m/AOC + m/COD All rights reserved. 42° and m/EOF= 66°. By the 180°. After application of the m/AOC 108°, and by the Reset Next DELL Feede ZEOF ZBOC. Therefore, m/COD Pienetas 72°
Transcribed Image Text:Prove: m/COD = 72° KA Complete the proof. It is given that m/AOB m/BOC = 66°. By the m/AOC + m/COD All rights reserved. 42° and m/EOF= 66°. By the 180°. After application of the m/AOC 108°, and by the Reset Next DELL Feede ZEOF ZBOC. Therefore, m/COD Pienetas 72°
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