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: 29


The angles of the triangle are as follows: $A = 80^{\circ}40', B = 64^{\circ}50',$ and $C = 34^{\circ}30'$ The lengths of the sides are as follows: $a = 156~cm, b = 143~cm,$ and $c = 89.6~cm$

Work Step by Step

We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{(143~cm)^2+(89.6~cm)^2-(2)(143~cm)(89.6~cm)~cos~80^{\circ}40'}$ $a = \sqrt{24321.4~cm^2}$ $a = 156~cm$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(156)^2+(89.6)^2-(143)^2}{(2)(156)(89.6)})$ $B = arccos(0.426)$ $B = 64^{\circ}50'$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-80^{\circ}40'-64^{\circ}50'$ $C = 34^{\circ}30'$
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