Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 1 - Trigonometric Functions - Section 1.2 Angle Relationships and Similar Triangles - 1.2 Exercises - Page 20: 74d


$CG~feet = EG~paces$

Work Step by Step

$\frac{CG~feet}{1~foot} = \frac{AG}{AD} = \frac{EG}{BD} = \frac{EG~paces}{1~pace}$ $\frac{CG~feet}{1} = \frac{EG~paces}{1}$ $CG~feet = EG~paces$
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