Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - Chapter Review - Review Exercises - Page 557: 26

Answer

$1.92$ square inches

Work Step by Step

The area of a sector can be computed using the formula: $K=\frac{\theta}{360}\cdot \pi \cdot r^2$,where $\theta$ is in degrees Hence, here we have $K_1=\frac{50}{360}\cdot \pi \cdot 6^2\approx15.71.$ By definition the area of a triangle is given by the formula: $K_2=\frac{ab\cdot \sin(C)}{2}.$ Hence, here we have $K_2=\frac{6\cdot6\cdot \sin(50^o)}{2}\approx13.79.$ Thus, the area of the segment is: $K_1-K_2=15.71-13.79=1.92$
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