Hello Sarah,
I would do part b) first, and then plot the resulting function after that.
Since the wheel is circular and rotates at presumably a constant rate, the height of the passengers will vary sinusoidally with time. Since we are given that they are at the minimum height at time $t=0$, then we need a negative cosine function.
Since the radius of the wheel is 135 m, this will be the amplitude, and the period of the motion is said to be 3 minutes, hence we may find the angular velocity $\omega$ by:
$$\frac{2\pi}{\omega}=3\,\therefore\,\omega=\frac{2\pi}{3}$$
The vertical translation is the radius plus 4 metres, or 139 metres, hence the height $h$ (in metres) as a function of time $t$ (in minutes) is given by:
b) $$h(t)=-135\cos\left(\frac{2\pi}{3}t \right)+139$$
a) Here is a plot of the function we found for part b) over the first period:
https://www.physicsforums.com/attachments/861._xfImport
c) To find the passengers' height at time $t=5$, we need to evaluate:
$$h(5)=-135\cos\left(\frac{2\pi}{3}5 \right)+139=-135\left(-\frac{1}{2} \right)=\frac{413}{2}=206.5$$
d) To find when the height is 85 m for the first time, we need to equate the height to 85 m and solve for the smallest positive time:
$$85=-135\cos\left(\frac{2\pi}{3}t \right)+139$$
Subtract through by 139, then multiply through by -1:
$$135\cos\left(\frac{2\pi}{3}t \right)=54$$
Divide through by 135 and reduce:
$$\cos\left(\frac{2\pi}{3}t \right)=\frac{2}{5}$$
The inverse cosine function will give us the value we want by default.
$$\frac{2\pi}{3}t=\cos^{-1}\left(\frac{2}{5} \right)$$
Solve for $t$ by multiplying through by $$\frac{3}{2\pi}$$
$$t=\frac{3}{2\pi}\cos^{-1}\left(\frac{2}{5} \right)\approx0.5535151793483181\approx0.6$$
To Sarah and any other guests viewing this topic, I invite and encourage you to post other trigonometric questions here in our http://www.mathhelpboards.com/f12/ forum.
Best Regards,
Mark.
