You work at a pioneer historical site. on this site you have handcarts. one cart has a handle that connects to the center of the wheel. we have to maintain the handle of the cart at an angle of no more than 20° with the ground so the contents do not spill out. the distance from where the handle rests on the ground to the point where the wheel is sitting on the ground is 45 inches. the distance of the center of the wheel to the end of the handle is approximately 48 inches.

a. if a pioneer pulling the handcart held the handle at a height of 48 inches off the ground, would the contents of the cart spill out the back? how high can the pioneer lift the handle off the ground before the contents started spilling out?

Let the center of the wheel be O, and let the endpoint of the handle as shown in the original picture be H.

Check the first picture attached. OH=48 and m(OHB)=20°.

|OB|=|OH|*sin20°=48 in * 0.342= 16.4 in

The handle of the handcart when lifted, moves around a circle formed, with center  and radius |OH|=48 in.

To elevate the handle at 48 in off the ground, the handle first comes at point H', such that OH'//BH, then comes to point H" such that the distance of point H" to the ground is 48 inches. (check picture 1)

check picture 2, the distance of the point H'' to the line segment OH' is 48-16.4= 31.6 (inches) 

31.6 = 48 * sinα

sin α=31.6/48=0.6583

so α = arcsin(0.6583)=41.17°

similarly, instead of α, 20° is the greatest angle such that the content dont spill out, and let h be the height above the line segment OH'.

h= 48*sin20=48*0.342=16.4

so the maximal height at which the content dont spill is 16.4+16.4=32.8 (inches)

1. the contents would spill out,  2. maximal height: 32.8 inches