Answer: quadrant III
Step-by-step explanation: srry if am wrong
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:
Step-by-step explanation:
h^2=x^2+y^2, since this is a square x=y=s so
220^2=2s^2
2s^2=48400
s^2=24200
s=24200^(1/2)m (exact)
s=155.56m (rounded to nearest hundredth of a meter)
One hectare is equal to 10000m^2
A=24200m^2(h/10000m^2)
A=2.42 hectares
C=(-10;5). If you want an explanation I would be happy to explain it
