Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - 6.1 Radian Measures - 6.1 Exercises - Page 574: 80


$43^o N$.

Work Step by Step

RECALL: The length of the arc $(s)$ intercepted by the central angle $\theta$ in a circle of radius $r$ is given by the formula: $s = r\theta$, where $\theta$ is in radian measure. The radius of Earth is around $6400$ km., and the distance between the two cities (which is $s$) is 1100 km. Use the formula above to obtain: $s=r\theta \\1100=6400 \cdot \theta \\\dfrac{1100}{6400} = \theta \\0.171875=\theta \\0.171875 \cdot \dfrac{180^o}{\pi}=\theta \\9.847712104^o=\theta \\\theta \approx 10^o$ Since Charleston is found below Toronto on the map, then the latitude of Toronto must be 10 degrees higher than Charleston. Thus, the latitude of Toronto must be: $=33^o+10^o \\=43^o N$.
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