Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 11 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 520: 21

Answer

γ = 55.1°

Work Step by Step

Using theorem 11.4.2 Law of sines: $\frac{sine α}{a}$ = $\frac{sine γ}{b}$ $\frac{sine 80°}{12 in}$ = $\frac{sine γ}{10 in}$ sine γ = $\frac{sine 80°}{12 in}$ * 10 in sine γ = 0.8206 γ = $sin^{-1}$(0.8206) γ = 55.1°
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