Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 11 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 528: 32

Answer

$α = 90^{\circ}$

Work Step by Step

1. Use the Law of Cosines to find $α$ $c^{2} = a^{2} + b^{2} - 2abcos(C)$ $(26)^{2} = (24)^{2} + (10)^{2} - 2(24)(10)cos(α)$ $676 = 576 + 100 - 480cos(α)$ $676 = 676 - 480cos(α)$ $0 = - 480cos(α)$ $cos(α) = 0$ $α = 90^{\circ}$
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