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 310: 14

Answer

The angles are $A=72.2^{\circ}, B=48.2^{\circ}$, and $C=59.6^{\circ}$ The angles are $A=107.8^{\circ}, B=48.2^{\circ}$, and $C=24.0^{\circ}$

Work Step by Step

We can use the law of sines to find the angle $A$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~A = \frac{a~sin~B}{b}$ $sin~A = \frac{(890~cm)~sin~(48.2^{\circ})}{697~cm}$ $A = arcsin(0.9519)$ $A = 72.2^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-72.2^{\circ}-48.2^{\circ}$ $C = 59.6^{\circ}$ The angles are $A=72.2^{\circ}, B=48.2^{\circ}$, and $C=59.6^{\circ}$ Note that we can also construct another triangle. $A = 180-72.2^{\circ} = 107.8^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-48.2^{\circ}-107.8^{\circ}$ $C = 24.0^{\circ}$ The angles are $A=107.8^{\circ}, B=48.2^{\circ}$, and $C=24.0^{\circ}$
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