Awheel with a dot on its edge rolls on the ground. the radius of the wheel is 15 inches. when the dot is at the position shown below, at an angle of 112°, what is the distance of the dot above the ground, to the nearest tenth of an inch?

The distance of the dot above the ground is 28.9 in

Step-by-step explanation:

see the attached figure with letters to better understand the problem

we know that

The triangle ABC is an isosceles triangle

CA=CB=15 in ------> is the radius of the circle

∠ACD+112°=180° ---> because the diameter divide the circle into two equal parts

∠ACD=180° -112°=68°

In the right triangle ACD

Find AD we have sin(68°)=AD/AC AD=AC*sin(68°) substitute the value AD=15*sin(68°)=13.9 in

Find the distance AB

AB=2*AD AB=2*13.9=27.8 in

The diameter is equal to

2*15=30 in

The distance of the dot above the ground is equal to

AB+(30-27.8)/2

27.8+1.1=28.9 in

all it means is function 3