Question

A person gets on a Ferris wheel at the starting point. The starting point is 15 feet off the ground. The Ferris wheel has a radius of 50 feet. What is the height of the rider from the ground after the Ferris wheel rotates 11π/12 radians?

Answer 1

Answer:

$\boxed{\frac{25}{2}(\sqrt{6}+\sqrt{2})+65}$

Explanation:

The height of the rider from the ground after the Ferris wheel equals the starting point is 15 feet off the ground plus the radius of the Ferris wheel plus the opposite side of the triangle ΔABC (See figure below). Hence our goal is to find this side. We have the angle $11\pi /2=165^{\circ}$, therefore the angle ∠BAC = 165° - 90° = 75°. So this side can be found using trigonometry:

$\overline{BC}=\overline{AB}sin(75^{\circ})=50sin(75^{\circ})=\frac{25}{2}(\sqrt{6}+\sqrt{2})$

Finally, the height is:

$H=\frac{25}{2}(\sqrt{6}+\sqrt{2})+15+50 \\ \\ \boxed{H=\frac{25}{2}(\sqrt{6}+\sqrt{2})+65}$