Elementary Geometry for College Students (6th Edition)

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

Chapter 11 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 519: 9b

Answer

γ = 90°

Work Step by Step

By using theorem 11.4.3 $c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab cos γ $5^{2}$ = $3^{2}$ + $4^{2}$ - 2*3*4* cos γ 25 = 9 + 16 - 24cos γ 25 = 25 - 24cos γ 25 - 25 = - 24cos γ 0 = - 24cos γ cos γ = 0 γ = $cos^{-1}$(0) γ = 90°
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