It is relatively simple to solve this problem by drawing. If CB is perpendicular to OA at point B, you will get a right triangle OBC. Where the angle BOC is the angle A, then H can be regarded as the distance from O to the ground, that is, 4.8+0.8=5.6, minus the AC length of the right angle, that is, 4.8*cosa. So there is H=5.6-4.8*cosa.

The solution of the second problem is similar to the first problem, that is, the two lengths are subtracted. It took 60 seconds to turn once, and now it has turned t seconds, so the radian number of angle A is calculated as t/60*2*π. Other conditions remain unchanged, so there is; h=5.6-4.8*cos(2*π*t/60)。 Note that the unit here is radian, and the first question is angle.