Trigonometry (11th Edition) Clone

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

Chapter 7 - Review Exercises - Page 349: 13


55.5 m

Work Step by Step

First, we convert minutes into decimal degrees: $51^{\circ}20'=51\frac{20}{60}=51.33^{\circ}$ We can use the law of cosines here because we know the lengths of two sides of the triangle and the measure of the included angle. The law of cosines is: $a^{2}=b^{2}+c^{2}-2bc\cos A$ where $b,c$ are the two known sides of the triangle while $A$ is the known angle. The unknown side opposite the known angle is $a$. Substituting the values in the formula and solving: $a^{2}=b^{2}+c^{2}-2bc\cos A$ $a^{2}=58.2^{2}+68.3^{2}-2(58.2)(68.3)\cos 51.33$ $a^{2}=3387.24+4664.89-7950.12\cos 51.33$ $a^{2}=8052.13-7950.12\cos 51.33$ Using a calculator, we find that $\cos 51.33^{\circ}=0.62483$. Therefore, $a^{2}=8052.13-7950.12\cos 51.33$ $a^{2}=8052.13-7950.12(0.62483)$ $a^{2}=8052.13-4967.50$ $a^{2}=3084.63$ $a=\sqrt {3084.63}$ $a=55.54\approx55.5$ Therefore, the length of the unknown side of the triangle is 55.5 m.
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