Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.6 - The Law of Cosines - 6.6 Exercises - Page 522: 48

Answer

$θ\approx31.17º$

Work Step by Step

The angle we can to find opposite to is the one that is 120ft long, so: $120^2=212^2+230^2-2(212)(230)\cdot $cos$(θ)$ $14'400=44'944+52'900-97'520\cdot $cos$(θ)$ $14'400-44'944-52'900=-97'520\cdot $cos$(θ)$ $14'400-44'944-52'900=-97'520\cdot $cos$(θ)$ $-83'444=-97'520\cdot $cos$(θ)$ cos$(θ)=\frac{-83'444}{-97'520}$ $θ=$cos$^{-1}\frac{-83'444}{-97'520}$ $θ\approx31.17º$
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