What mistake did I make while solving the Ferris wheel trig problem?

[NeedHelp27]

Hello, all. For homework, we got a problem that reads as follows: A Ferris wheel 50 ft in diameter makes one revolution every 40 sec. If the center of the wheel is 30 ft above the ground, how long after reaching the low point is a rider 50 ft above the ground? Our teacher said to model the situation with an equation.

When I tried to go about doing this, I drew a graph showing the person's height, at it ended up being the cosine graph shifted up 5 and over so that a low point was on the y-axis (0,5). Next, I tried to write the equation in the form f(x)=acos(bx+c)+d. I did (max value - min value)/2 = (55-5)/2=25 to find a. Then, because one revolution takes 40 seconds, I solved 2pi/b=40 for b and got b=pi/20. The graph is shifted 5 up, so d-5. That gave me f(x)=25cos((pix)/20+c)+5. I also have the point (0,5) on the graph, so I can plug it into get c. f(0)=5=25cos(c)+5. C ended up being pi/2.

Then I went to solve the problem.
f(t)=50=25cos((pix)/20+pi/2)+5.
When I finally isolated x, I got a domain error. Where did I go wrong?

 

[MarkFL]

If we let the point in time (in seconds) where the rider is at the lowest point be $t=0$, then we could write:

\(\displaystyle f(t)=-25\cos\left(\frac{\pi}{20}t\right)+30\)

Now try solving the question. :D