## Precalculus (6th Edition) Blitzer

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