Ferris wheel problem

Question

A Ferris wheel is boarding platform is 4 meters above the ground, has a diameter of 72 meters, and rotates once every 10 minutes.

How many minutes of the ride are spent higher than 48 meters above the ground?

Round to three decimal places.

Answer 1

48 meters above ground is 4 m above center and the angle above center arcsin(4/36)=.0668 radians. The angle of rotation above 48m = π-2*.0668= 3 radians. The angular velocity = 2π/10 rad/min. So, the amount of time above 48m = 3*5/π= 4.79 min

Answer 2

the approach by Stephen is accurate, but angle that is created is sin^-1(8/36)

much agree must use (π-2sin^-1(8/36))(10/2π)=4.286

which is same as Robert

Answer 3

I am getting a slightly different answer writing as a cosine function such that one 10 minute cycle starts and ends at the bottom.

H(t) = -36 Cos (pi/5)t + 40

I get the pi/5 from B = (2pi)/per = (2pi)/10 = pi/5

Now putting in 48 for H(t) I get

8 = -36 Cos (pi/5)t

simplified further to

Cos(pi/5)t = - 2/9

NowI get (pi/5)/t = 1.794 directly on calculator as 2nd quadrant answer -- for on the way up.

and also need (pi/5)t = 4.488 as 3rd quadrant answer -- for on the way down.

Solving both for t I get t = 2.857 and t = 7.143

The difference is 4.286 minutes