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: 77

Answer

$135^{\circ}$

Work Step by Step

Step 1: We let $\textbf {u}=\langle 2,1 \rangle$ and $\textbf {v}=\langle -3,1 \rangle$ 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: $\cos\theta=\frac{\langle 2,1 \rangle\cdot\langle -3,1 \rangle}{|\langle 2,1 \rangle||\langle -3,1 \rangle|}$ Step 4: $\cos\theta=\frac{2(-3)+1(1)}{\sqrt (2^{2}+1^{2})\cdot\sqrt ((-3)^{2}+1^{2})}$ Step 5: $\cos\theta=\frac{-6+1}{\sqrt (4+1)\cdot\sqrt (9+1)}$ Step 6: $\cos\theta=\frac{-5}{\sqrt (5)\cdot\sqrt (10)}$ Step 7: $\cos\theta=\frac{-5}{\sqrt (5)\sqrt (2)\sqrt (5)}$ Step 8: $\cos\theta=\frac{-5}{5\sqrt 2}$ Step 9: $\cos\theta=\frac{-1}{\sqrt 2}$ Step 10: $\theta=\cos^{-1}(\frac{-1}{\sqrt 2})$ Step 11: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(\frac{-1}{\sqrt 2})=135^{\circ}$
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