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: 22


There is one triangle that exists with the given parts. $A = 25.5^{\circ}, B = 102.2^{\circ}$, and $C = 52.3^{\circ}$ $a = 32.5~yd, b = 73.9~yd$, and $c = 59.8~yd$

Work Step by Step

We can use the law of sines to find the angle $A$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $sin~A = \frac{a~sin~C}{c}$ $sin~A = \frac{(32.5~yd)~sin~(52.3^{\circ})}{59.8~yd}$ $A = arcsin(0.43)$ $A = 25.5^{\circ}$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-25.5^{\circ}-52.3^{\circ}$ $B = 102.2^{\circ}$ We can find the length of side $b$: $\frac{b}{sin~B} = \frac{c}{sin~C}$ $b = \frac{c~sin~B}{sin~C}$ $b = \frac{(59.8~yd)~sin~(102.2^{\circ})}{sin~52.3^{\circ}}$ $b = 73.9~yd$ Note that we can also find another angle for A. $A = 180-25.5^{\circ} = 154.5^{\circ}$ However, we can not form a triangle with this angle A and angle C since these two angles sum to more than $180^{\circ}$
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