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: 21


There is one possible triangle with the given parts. $A = 42.5^{\circ}, B = 20.6^{\circ},$ and $C = 116.9^{\circ}$ $a = 15.6~ft, b = 8.14~ft,$ and $c = 20.6~ft$

Work Step by Step

We can use the law of sines to find the angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(8.14~ft)~sin~(42.5^{\circ})}{15.6~ft}$ $B = arcsin(0.3525)$ $B = 20.6^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-42.5^{\circ}-20.6^{\circ}$ $C = 116.9^{\circ}$ We can find the length of side $c$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(15.6~ft)~sin~(116.9^{\circ})}{sin~42.5^{\circ}}$ $c = 20.6~ft$ Note that we can also find another angle for B. $B = 180-20.6^{\circ} = 159.4^{\circ}$ However, we can not form a triangle with this angle B and angle A since these two angles sum to more than $180^{\circ}$
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