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°

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.1: Angles
Problem 21E
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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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