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: 51


$46.4$ m$^{2}$

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: $Area=\frac{1}{2}bc \sin A$ We substitute the values of $A,b$ and $c$ in this formula and solve: $Area=\frac{1}{2}bc \sin A$ $Area=\frac{1}{2}(13.6)(10.1) \sin 42.5^{\circ}$ $Area=\frac{1}{2}(137.36) \sin 42.5^{\circ}$ $Area=68.68 \sin 42.5^{\circ}$ Using a calculator, $\sin 42.5^{\circ}=0.6756$. Therefore, $Area=68.68 \sin 42.5^{\circ}$ $Area=68.68(0.6756)$ $Area=46.400=46.4$ Therefore, the area of the triangle is $46.4$ m$^{2}$.
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