Aconstruction crew wants to hoist a heavy beam so that it is standing up straight. ey tie a rope to the beam, secure the base, and pull the rope through a pulley to raise one end of the beam from the ground. when the beam makes an angle of 408 with the ground, the top of the beam is 8 ft above the ground. e construction site has some telephone wires crossing it. e workers are concerned that the beam may hit the wires. when the beam makes an angle of 608 with the ground, the wires are 2 ft above the top of the beam. will the beam clear the wires on its way to standing up straight? explain.

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?