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 upslope and one directly downslope. 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 90A 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(90A)
Note; cos(90A)=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 20140618 20:48:18 