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 $17\text{ feet}$.
