Hello, topaz192!
Domingo decides to ride the ferris wheel at the carnival.
When he gets into a seat at the bottom of the ferris wheel, he is 4 ft above the ground.
a) If the radius of the ferris wheel is 36 feet,
how far above the ground will Domingo be when his seat reaches the top? You really can't figure this out ??
The wheel has a radius of 36 feet  a diameter of 72 feet.
He starts 4 feet above the ground.
. . At the top, he has risen 72 feet more.
Therefore . . .
b) The ferris wheel rotates 300° CCW and stops.
How far above the ground is Domingo when it stops? Code:
* * *
* *
* *
* *
* O *
* o *
* 36 *  *
*60°
* * B *
Ao    o    *
*  *
* o *
C
4

      o      
D
The center of the ferris wheel is $\displaystyle O.$
It radius is: $\displaystyle OA = 36$ feet. .$\displaystyle OD = 40$ feet.
Angle AOB = 60°
$\displaystyle BD\:=\:40  OB$ is Domingo's height above the ground.
Triangle OBA is a 3060 right triangle.
. . Hence: .$\displaystyle OB \:= \:\frac{\text{radius}}{2} \:=\:18$ feet.
Therefore, his height is: .$\displaystyle BD \:=\:40  18 \:=\:22$ feet.
c) Suppose the radius of the ferris wheel is only 30 ft.
How far above the ground is Domingo after the ferris wheel rotates 300° ? If the radius is 30, then: .$\displaystyle OD\:=\:34$
$\displaystyle BD \:=\:34  OB$
$\displaystyle OB \:=\:\frac{30}{2} \:=\:15$ feet.
Therefore, his height is: .$\displaystyle BD \:=\:34  15 \:=\:19$ feet.
d) Suppose the radius of the ferris wheel is $\displaystyle R$ feet.
Write an expression for Domingo's height after the ferris wheel rotates 300°. We have: .$\displaystyle OD \:=\:R + 4$ feet.
$\displaystyle OB \:=\:\frac{R}{2}$ feet.
Therefore, his height is: .$\displaystyle BD \;=\;(R + 4)  \frac{R}{2} \;=\;\frac{R}{2} + 4$ feet.