Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Mid-Chapter Check Point - Page 756: 22


See below:

Work Step by Step

We are given $\theta =\frac{\pi }{3}$. Therefore, $\begin{align} & \tan \theta =\tan \frac{\pi }{3} \\ & \tan \theta =\sqrt{3} \end{align}$ As $ x=r\cos \theta \ \text{ and }\ y=r\sin \theta $. $\begin{align} & \frac{r\sin \theta }{r\cos \theta \ }=\frac{y}{x} \\ & \tan \theta =\frac{y}{x} \end{align}$ Also, $\tan \theta =\sqrt{3}$ Therefore, $\begin{align} & \frac{y}{x}=\sqrt{3} \\ & y=\sqrt{3}x \end{align}$ Thus, the provided equation is the equation of line with slope $\sqrt{3}$.
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.