Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises: 21

Answer

The angles of the triangle are as follows: $A = 53.1^{\circ}, B = 81.3^{\circ},$ and $C = 45.6^{\circ}$ The lengths of the sides are as follows: $a = 7.23~m, b = 8.94~m,$ and $c = 6.46~m$

Work Step by Step

$a = 7.23~m$ $b = 8.94~m$ $C = 45.6^{\circ}$ We can use the law of cosines to find $c$: $c^2 = a^2+b^2-2ab~cos~C$ $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(7.23~m)^2+(8.94~m)^2-(2)(7.23~m)(8.94~m)~cos~45.6^{\circ}}$ $c = \sqrt{41.749~m^2}$ $c = 6.46~m$ We can use the law of sines to find $A$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~C}{c}$ $A = arcsin(\frac{a~sin~C}{c})$ $A = arcsin(\frac{7.23~sin(45.6^{\circ})}{6.46})$ $A = arcsin(0.799634)$ $A = 53.1^{\circ}$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-53.1^{\circ}-45.6^{\circ}$ $B = 81.3^{\circ}$
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