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: 33


The angles of the triangle are as follows: $A = 112.8^{\circ}, B = 21.7^{\circ},$ and $C = 45.5^{\circ}$ The lengths of the sides are as follows: $a = 15.7~m, b = 6.28~m,$ and $c = 12.2~m$

Work Step by Step

We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{(6.28~m)^2+(12.2~m)^2-(2)(6.28~m)(12.2~m)~cos~112.8^{\circ}}$ $a = \sqrt{247.658~m^2}$ $a = 15.7~m$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(15.7)^2+(12.2)^2-(6.28)^2}{(2)(15.7)(12.2)})$ $B = arccos(0.929)$ $B = 21.7^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-112.8^{\circ}-21.7^{\circ}$ $C = 45.5^{\circ}$
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