Interesting problem ...
The key is to realize that the wires have some distance to the ground, that does not change.
The pole does change. But the vertical height of the pole plus the distance from the pole to the wires is the distance ground to the wires all the time. In other words, for any angle one has:
D = L * sin(alpha) + d, where D is the distance wires-ground, L is the length of the pole, alpha is the angle, and 'd' is the distance from the top of the (inclined) pole to the wires:
L*sin(40) + 8 = L*sin(60) + 2, so one can get the length of the pole:
L = (8-2)/(sin(60) - sin(40)) = 6/0.2232 = 26.88 ft (be careful to have the calculator in degrees not rad)
So the pole is 26.88 ft long!
If the wires are higher than 26.88 ft, no problem. if they are below, the concerns are justified and it won't pass!
Your statement does not mention the distance between the wires and the ground. Do you have it?
Answer:
Mine has been good so far
Step-by-step explanation:
I believe 9+10=21, but I’m not sure.
Parameterize the surface (call it
) by
with
and
. Then the surface element is
The area of
is then given by the surface integral
if you convert it all into one hour, Team B clearly worked at a faster pace.
