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: 42


$A = 40.41~cm^2$

Work Step by Step

Since the vertex angle is $46^{\circ}$, we can let $\theta = \frac{46^{\circ}}{2} = 23^{\circ}$ We can find the area of the triangle: $A = s^2~sin~\theta~cos~\theta$ $A = (10.6~cm)^2~sin~23^{\circ}~cos~23^{\circ}$ $A = 40.41~cm^2$
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