Trigonometry (10th Edition)

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 305: 31

Answer

$sin~C = 1$ $C = 90^{\circ}$ Triangle ABC is a $30^{\circ},60^{\circ},90^{\circ}$ triangle.

Work Step by Step

We can use the law of sines to find the angle $C$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $sin~C = \frac{c~sin~A}{a}$ $sin~C = \frac{(2\sqrt{5})~sin~(30^{\circ})}{\sqrt{5}}$ $sin~C = 1$ $C = arcsin(1)$ $C = 90^{\circ}$ Triangle ABC is a $30^{\circ},60^{\circ},90^{\circ}$ triangle.
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