Trigonometry (10th Edition)

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

Chapter 7 - Test - Page 347: 12

Answer

$41.8^{\circ}$

Work Step by Step

Step 1: We substitute vectors $\textbf {u}$ and $\textbf {v}$ in the formula for finding the angle between a pair of vectors, $\cos\theta=\frac{\textbf {u}\cdot\textbf {v}}{|\textbf {u}||\textbf {v}|}$ Step 2: $\cos\theta=\frac{\langle 4,3 \rangle\cdot\langle 1,5 \rangle}{|\langle 4,3 \rangle||\langle 1,5 \rangle|}$ Step 3: $\cos\theta=\frac{4(1)+3(5)}{\sqrt (4^{2}+3^{2})\cdot\sqrt (1^{2}+5^{2})}$ Step 4: $\cos\theta=\frac{4+15}{\sqrt (16+9)\cdot\sqrt (1+25)}$ Step 5: $\cos\theta=\frac{19}{\sqrt (25)\cdot\sqrt (26)}$ Step 6: $\cos\theta=\frac{19}{5\times\sqrt (26)}$ Step 7: $\cos\theta=\frac{19}{5\sqrt (26)}$ Step 8: $\theta=\cos^{-1}(\frac{19}{5\sqrt (26)})$ Step 9: Solving using the inverse cos function on the calculator, $\theta=\cos^{-1}(\frac{19}{5\sqrt (26)})\approx41.8^{\circ}$
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