Trigonometry (10th Edition)

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

Chapter 2 - Quiz (Sections 2.1-2.3) - Page 68: 8

Answer

$\theta = 60^{\circ}$ $\theta = 120^{\circ}$

Work Step by Step

$sin~\theta = \frac{\sqrt{3}}{2}$ $\frac{y}{r} = \frac{\sqrt{3}}{2}$ We can let $y = \sqrt{3}$ and $r = 2$. Then: $x^2 = r^2-y^2$ $x = \pm \sqrt{r^2-y^2}$ $x = \pm \sqrt{(2)^2-(\sqrt{3})^2}$ $x = 1$ or $x = -1$ If $x = 1$ and $y = \sqrt{3}$, then $\theta$ makes an angle of $60^{\circ}$ above the positive x-axis. Then $\theta = 60^{\circ}$ If $x = -1$ and $y = \sqrt{3}$, then $\theta$ makes an angle of $60^{\circ}$ above the negative x-axis. Then $\theta = 180^{\circ}-60^{\circ} = 120^{\circ}$
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.