Trigonometry (10th Edition)

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.4 Vectors, Operations, and the Dot Product - 7.4 Exercises - Page 331: 80

Answer

$45^{\circ}$

Work Step by Step

Step 1: We let $\textbf {u}=\langle 4,0 \rangle$ and $\textbf {v}=\langle 2,2 \rangle$ Step 2: The formula for finding the angle between a pair of vectors is $\cos\theta=\frac{\textbf {u}\cdot\textbf {v}}{|\textbf {u}||\textbf {v}|}$ Step 3: $\cos\theta=\frac{\langle 4,0 \rangle\cdot\langle 2,2 \rangle}{|\langle 4,0 \rangle||\langle 2,2 \rangle|}$ Step 4: $\cos\theta=\frac{4(2)+0(2)}{\sqrt (4^{2}+0^{2})\cdot\sqrt (2^{2}+2^{2})}$ Step 5: $\cos\theta=\frac{8+0}{\sqrt (16+0)\cdot\sqrt (4+4)}$ Step 6: $\cos\theta=\frac{8}{\sqrt (16)\cdot\sqrt (8)}$ Step 7: $\cos\theta=\frac{8}{4\times\sqrt (4\times2)}$ Step 8: $\cos\theta=\frac{8}{4\times\sqrt 4\times\sqrt 2}$ Step 9: $\cos\theta=\frac{8}{8\sqrt 2}$ Step 10: $\cos\theta=\frac{1}{\sqrt 2}$ Step 11: $\theta=\cos^{-1}(\frac{1}{\sqrt 2})$ Step 12: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(\frac{1}{\sqrt 2})=45^{\circ}$
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