math trigonnometry hrlp?

Question

i need help on this problem?
the eye if a basketball player are 6feet 2 inches above the floor. the player is at the free-throw line, which is 15 feet from the center of the basketball rim. what is the angle of elevation from the player eyes to the center of the rim?
included picture would be great...

Answer 1

The rim is 10 ft off the floor, right? So eye level to rim vertical distance is 10 - 6 2/12 = 3 10/12 ft. If angle of elevation is A,

tan A = (3 5/6) ÷ 15 = 23/90, so
A = inv tan 23/90 = 14.335°

Answer 2

How high is the rim?