Question 1079524
We want to know the fraction of the wheel's edge that is more than 9 meters above the ground.
{{{drawing(300,300,-7.5,7.5,-2,13,
circle(0,6,5),circle(0,6,0.2),
line(8,0,-8,0),green(arrow(-8,9,12,9)),
rectangle(-1,0,1,6.5),green(triangle(0,6,0,9,4,9)),
green(rectangle(0,9,0.3,8.7)),
green(arrow(-7,0,-7,9)),green(arrow(-7,9,-7,0)),
locate(-6.8,4,9m),locate(0.2,4,6m),
red(arrow(0,0,0,6)),red(arrow(0,6,0,0)),
green(triangle(0,6,0,9,-4,9)),locate(-0.2,9.8,B),
locate(4.1,9.6,P),locate(-4.5,9.7,Q),locate(0.3,6.3,A)
)}}} The small arc between points P and Q represents the part of the ride when Jamie sees the ocean.
{{{AB=9m-6m=3m}}} , and {{{AP=10m/2=5m}}} is the radius of the wheel.
So, in right triangle BAP,
{{{cos(BAP)=AB/AP=3/5=0.6}}} ---> {{{BAP=about53.13^o}}} (approximately
{{{PAQ=about}}}{{{2*53.13^o=106.26^o}}}
The fraction of each revolution that Jamie is between P and Q is
{{{106.26^o/360^o=106.26/360}}} .
{{{5 minutes=5*60seconds}}}, so the time that Jamie sees the ocean is
{{{5*60seconds(106.26^o/360^o)=88.55seconds}}}