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 527: 4b

Answer

$\frac{sine 40°}{5.3}$ = $\frac{sine 60°}{c}$

Work Step by Step

Given a=5.3, α =40° β=80° Using theorem 11.4.2 By law of sines $\frac{sine α}{a}$ = $\frac{sine γ}{c}$ $\frac{sine 40°}{5.3}$ = $\frac{sine (180° -(α + β))}{c}$ $\frac{sine 40°}{5.3}$ = $\frac{sine (180° -(40 + 80))}{c}$ $\frac{sine 40°}{5.3}$ = $\frac{sine (180° -120)}{c}$ $\frac{sine 40°}{5.3}$ = $\frac{sine 60°}{c}$
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