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 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 311: 28


There is one possible triangle with the given parts. $A = 60.91^{\circ}, B = 30.39^{\circ},$ and $C = 88.70^{\circ}$ $a = 98.25~m, b = 56.87~m,$ and $c = 112.4~m$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $sin~B = \frac{(56.87~m)~sin~(88.70^{\circ})}{112.4~m}$ $B = arcsin(0.5058)$ $B = 30.39^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-30.39^{\circ}-88.70^{\circ}$ $A = 60.91^{\circ}$ We can find the length of side $a$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $a = \frac{c~sin~A}{sin~C}$ $a = \frac{(112.4~m)~sin~(60.91^{\circ})}{sin~88.70^{\circ}}$ $a = 98.25~m$ Note that we can also find another angle for B. $B = 180-30.39^{\circ} = 149.61^{\circ}$ However, we can not form a triangle with this angle B and angle C since these two angles sum to more than $180^{\circ}$
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