## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Review Exercises - Page 352: 44

#### Answer

$142.1^{\circ}$

#### Work Step by Step

Step 1: We let $\textbf {u}=\langle 3,-2 \rangle$ and $\textbf {v}=\langle -1,3 \rangle$ 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: $\cos\theta=\frac{\langle 3,-2 \rangle\cdot\langle -1,3 \rangle}{|\langle 3,-2 \rangle||\langle -1,3 \rangle|}$ Step 4: $\cos\theta=\frac{3(-1)-2(3)}{\sqrt (3^{2}+(-2)^{2})\cdot\sqrt ((-1)^{2}+3^{2})}$ Step 5: $\cos\theta=\frac{-3-6}{\sqrt (9+4)\cdot\sqrt (1+9)}$ Step 6: $\cos\theta=\frac{-9}{\sqrt (13)\cdot\sqrt (10)}$ Step 7: $\cos\theta=\frac{-9}{\sqrt (130)}$ Step 8: $\theta=\cos^{-1}(\frac{-9}{\sqrt (130)})$ Step 9: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(\frac{-9}{\sqrt (130)})\approx142.1^{\circ}$ The vectors are not orthogonal.

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