Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 11 - Section 11.2 - The Cosine Ratio and Applications - Exercises - Page 504: 43


$A = 6.77~in^2$

Work Step by Step

Since the measure of a base angle is $72^{\circ}$, we can let $\theta = 180^{\circ}-72^{\circ}-90^{\circ} = 18^{\circ}$ We can find the area of the triangle: $A = s^2~sin~\theta~cos~\theta$ $A = (4.8~in)^2~sin~18^{\circ}~cos~18^{\circ}$ $A = 6.77~in^2$
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