Answer
$(1,0)$
Work Step by Step
Here, $r=\sqrt {x^2+y^2}=\sqrt{(1)^2+(0)^2}= 1$
$\tan \theta =\dfrac{y}{x}$
and $\theta=arctan[\dfrac{0}{1}]$
Thus, $\theta =0$
Hence, $(1,0)$
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 6 - Review Exercises - Page 798: 36
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