Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Review Exercises - Page 343: 22

Answer

$20.3$ ft$^{2}$

Work Step by Step

First, we convert decimal minutes to degrees: $35^{\circ}10'=35\frac{10}{60}=35.17^{\circ}$ The area of the triangle is half the product of the length of two sides and the sine of the angle included between them: $Area=\frac{1}{2}ab \sin C$ We substitute the values of $a,b$ and $C$ in this formula and solve: $Area=\frac{1}{2}ab \sin C$ $Area=\frac{1}{2}(6.90)(10.2) \sin 35.17^{\circ}$ $Area=\frac{1}{2}(70.38) \sin 35.17^{\circ}$ $Area=35.19\sin 35.17^{\circ}$ Using a calculator, $\sin 35.17^{\circ}=0.57600$. Therefore, $Area=35.19\sin 35.17^{\circ}$ $Area=35.19(0.57600)$ $Area=20.27$ Therefore, the area of the triangle is $20.3$ ft$^{2}$.
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