A tall tower is constructed on the side of a hill with uniform incline. Two guy wires are attached to the top of the tower and to points on the ground 50.0 feet from the tower - one directly up-slope and one directly down-slope. If the wires are 284.7 feet and 303.5 feet respectively, what is the angle of incline of the hill and how high is the tower?

If the slope is angle A to the horizontal, then the angle of the tower to the slope is 90-A on the uphill side and 90+A on the downhill side. (Tower is assumed vertical). Tower height is "h".

Consider the two triangles formed by the tower, wires and slope.  Using the law of cosines:
Uphill side: 284.7^2=h^2+50^+2*h*50*cos(90-A)
Note; cos(90-A)=sin(A)
Downhill side: 303.5^2=h^2+50^+2*h*50*cos(90+A) Note: cos(90+A)=-sin (A)

Adding the two equations eliminates A, and therefore h=288.97.

Plugging this value back into one of the cosine equations gives A = 10.99 deg

REF: Bractals' question - it does seem like the answer was intended to give integer results

Posted by Kenny M on 2014-06-18 20:48:18