Answer
$|v|=5$
$Θ=53°$
Work Step by Step
$V=\lt3,4\gt$
$=31+4j$
$|v|=\sqrt (3^{2} 4^{2})$
$=5$
$tanΘ=\frac{4}{3}$
$Θ=tan^{-1}=\frac{4}{3}$
$Θ=53°$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 9 - Section 9.1 - Vectors in Two Dimensions - 9.1 Exercises - Page 638: 47
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