Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 550: 104

Answer

The height of the seat from the ground after a $765{}^\circ $ rotation is $17\text{ feet}$.

Work Step by Step

Let us consider the diagram shown below: Here, O is the center of the wheel, and the distance from the center of the Ferris wheel to each seat is 40 feet. This means that $OA=OA'=40\text{ feet}$ Also, at the start of the ride, the height of the seat is 5 feet above the ground. Therefore, $AM=5\text{ feet}$ Now consider $\Delta OA'N$, where N is a right angle, and also consider $\angle A'ON=\theta $. $\begin{align} & \cos \theta =\frac{ON}{OA'} \\ & ON=OA'cos\theta \\ & =40\cos \theta \end{align}$ Put $765{}^\circ $ for $\theta $ to find out ON: $\begin{align} & ON=40\cos 765{}^\circ \\ & =40\cos \left( 2\times 360{}^\circ +45{}^\circ \right) \\ & =40\cos 45{}^\circ \end{align}$ At that instant, the height of the seat from the ground is $\begin{align} & A'M=NM \\ & =OM-ON \\ & =45-40\cos 45{}^\circ \\ & \approx 17\text{ feet} \end{align}$ Hence, the height of the seat from the ground after a $765{}^\circ $ rotation is $17\text{ feet}$.
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