## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Quiz (Sections 2.1-2.3) - Page 71: 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 $\theta = 180^{\circ}-60^{\circ} = 120^{\circ}$

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