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: 35


The angles of the triangle are as follows: $A = 29.9^{\circ}, B = 56.3^{\circ},$ and $C = 93.8^{\circ}$ The lengths of the sides are as follows: $a = 3.0~ft, b = 5.0~ft,$ and $c = 6.0~ft$

Work Step by Step

We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{3.0^2+6.0^2-5.0^2}{(2)(3.0)(6.0)})$ $B = arccos(\frac{5}{9})$ $B = 56.3^{\circ}$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{3.0^2+5.0^2-6.0^2}{(2)(3.0)(5.0)})$ $C = arccos(-\frac{1}{15})$ $C = 93.8^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-56.3^{\circ}-93.8^{\circ}$ $A = 29.9^{\circ}$
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