Trigonometry (10th Edition)

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

Chapter 7 - Review Exercises - Page 343: 19


The angles of the triangle are as follows: $A = 91^{\circ}50', B = 45^{\circ}40',$ and $C = 42^{\circ}30'$ The lengths of the sides are as follows: $a = 27.6~cm, b = 19.8~cm,$ and $c = 18.7~cm$

Work Step by Step

$a = 27.6~cm$ $b = 19.8~cm$ The angle $C = 42^{\circ}30'$ We can use the law of cosines to find $c$: $c = \sqrt{a^2+b^2-2ab~cos~C}$ $c = \sqrt{(27.6~cm)^2+(19.8~cm)^2-(2)(27.6~cm)(19.8~cm)~cos~42^{\circ}30'}$ $c = \sqrt{347.985~cm^2}$ $c = 18.7~cm$ We can use the law of sines to find angle $B$: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $B = arcsin(\frac{b~sin~C}{c})$ $B = arcsin(\frac{(19.8)~sin~42^{\circ}30'}{18.7})$ $B = arcsin(0.715)$ $B = 45^{\circ}40'$ We can find angle $A$: $A+B+C=180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-45^{\circ}40'-42^{\circ}30'$ $A = 91^{\circ}50'$
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