Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 7 - Applications of Trigonometric Functions - Section 7.4 Area of a Triangle - 7.4 Assess Your Understanding - Page 565: 34


$0.69 \ square \ inches$

Work Step by Step

The Area of sector $K_1$ can be determined by the formula: $K_1=\dfrac{\theta}{360}\cdot \pi \cdot r^2$, where $\theta$ is in degrees So, $K_1=\dfrac{40}{360} \ (\pi) (5^2) \approx 8.73$ The area of a triangle $K_2$ with sides $a,b$ and $c$ is given by: $K_2=\dfrac{1}{2} ab\ \sin(C)$ So, $K_2=\dfrac{(5)(5) \ \sin(40^{\circ})}{2}\approx 8.04 $ Therefore, the area of the segment is the difference between the areas of the sector and triangle: $K_1-K_2=8.73-8.04=0.69 \ square \ inches$
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