Elementary Geometry for College Students (6th Edition)

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

Chapter 11 - Section 11.1 - The Sine Ratio and Applications - Exercises - Page 497: 36c


$94.0 in^2$

Work Step by Step

The length of segment BC is: $ sin( 90 - \beta) = \frac{CB}{AB}$ $AB sin( 90 - \beta) = CB$ $CB=sin(90-55)(20 in)=11.5 in$ We now find the length of the other side: $sin( \beta) = AC/AB$ Simplifying, we find: $ABsin( \beta) = AC$ This gives: $AC=20(sin(55))=16.4 in$ Using the area formula: $A=.5bh$ $A=.5(16.4)(11.5)$ $A=94.0 in^2$
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