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?
Call the point where the down-slope guy wire touches the ground point A, and the up-slope guy wire touches the ground at point C. Point B is the top of the tower.
As usual, label the sides of triangle ABC with the lower-case version of the angle opposite.
a=284.7
b=100
c=303.5
a^2 = b^2 + c^2 -2*b*c*cosA
cosA=(b^2+c^2 - a^2) / (2*b*c)
Call the base of the tower D and the length of the tower itself t.
t^2 = c^2 + 50^2 - 2*c*50*cosA
The tower height t comes out to 289.97098... feet (let's say 290.0 feet, considering the precision of the given values).
Worked the other way for a check:
c^2 = a^2 + b^2 - 2*a*b*cosC
cosC = (a^2+b^2 - c^2) / (2*a*b)
t^2 = a^2 + 50^2 - 2*a*50*cosC
The obtuse angle at D can be found from:
sinD / c = sinA / t
sinD = c*sinA / t
D comes out to 180 - 79.0077... degrees, or 100.992296... degrees.
The acute angle at D:
sinD = a*sinC / t
with the same result, only this time we don't take the supplement.
The angle of incline of the hill is then the complement of 79.0077... degrees, or 90 less than 100.992296... degrees, or 10.992296... degrees (about 11°).
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Posted by Charlie
on 2014-06-19 09:57:57 |
solution
