Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 304: 54

Answer

$163$ km$^{2}$

Work Step by Step

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: $Area=\frac{1}{2}ab \sin C$ $Area=\frac{1}{2}(21.9)(24.6) \sin 142.7^{\circ}$ $Area=\frac{1}{2}(538.74) \sin 142.7^{\circ}$ $Area=269.37 \sin 142.7^{\circ}$ Using a calculator, $\sin 142.7^{\circ}=0.6060$. Therefore, $Area=269.37 \sin 142.7^{\circ}$ $Area=269.37(0.6060)$ $Area=163.24\approx163$ Therefore, the area of the triangle is $163$ km$^{2}$.
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