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: 52

Answer

$732$ ft$^{2}$

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: $Area=\frac{1}{2}ab \sin C$ $Area=\frac{1}{2}(35.1)(43.8) \sin 72.2^{\circ}$ $Area=\frac{1}{2}(1537.38) \sin 72.2^{\circ}$ $Area=768.69 \sin 72.2^{\circ}$ Using a calculator, $\sin 72.2^{\circ}=0.9521$. Therefore, $Area=768.69 \sin 72.2^{\circ}$ $Area=768.69(0.9521)$ $Area=731.89\approx732$ Therefore, the area of the triangle is $732$ ft$^{2}$.
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