Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises: 53

Answer

$r=15\sqrt {2};$ $p=15$ $t=10\sqrt {6}$ $q=5\sqrt {6}$

Work Step by Step

$r=\dfrac {15}{\cos 45}=15\sqrt {2};$ $p=r\sin 45=15$ $t=\dfrac {r}{\cos 30}=\dfrac {15\sqrt {2}}{\dfrac {\sqrt {3}}{2}}=10\sqrt {6}$ $q=r\tan 30^{0}=\dfrac {15\sqrt {2}}{\sqrt {3}}=5\sqrt {6}$
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.