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: 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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.