Precalculus (6th Edition)

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

Chapter 5 - Trigonometric Functions - 5.1 Angles - 5.1 Exercises: 2

Answer

$90^0$

Work Step by Step

In order to be complement $x+\theta =90^0\Rightarrow \theta =90^0-x^{0}=\dfrac {\pi }{2}$
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.