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 7.3 The Law of Cosines - 7.3 Exercises - Page 320: 32


The angles of the triangle are as follows: $A = 48.8^{\circ}, B = 71.5^{\circ},$ and $C = 59.7^{\circ}$ The lengths of the sides are as follows: $a = 3.73~mi, b = 4.70~mi,$ and $c = 4.28~mi$

Work Step by Step

We can use the law of cosines to find $c$: $c^2 = a^2+b^2-2ab~cos~C$ $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(3.73~mi)^2+(4.70~mi)^2-(2)(3.73~mi)(4.70~mi)~cos~59.7^{\circ}}$ $c = \sqrt{18.313~mi^2}$ $c = 4.28~mi$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(3.73)^2+(4.28)^2-(4.70)^2}{(2)(3.73)(4.28)})$ $B = arccos(0.3176)$ $B = 71.5^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-71.5^{\circ}-59.7^{\circ}$ $A = 48.8^{\circ}$
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