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: 24

Answer

β = 90°

Work Step by Step

using theorem 11.4.3 Law of cosine a=6,b =10, c=8 $b^{2}$ = $a^{2}$ + $c^{2}$ - 2ac cos β $10^{2}$ = $(6)^{2}$ + $(8)^{2}$ - 2*6*8*cos β 100 = 36+64 - 96cos β 96cos β = 36 +64 -100 96cos β = 0 cos β = 0 β = $cos^{-1}$(0) β = 90°
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