Asked by Anonymous
a tower that is a 200 meters is leaning to one side. from a certain point on that side, the angle of elevation to the top of the tower is 70 degree. From a point 55 meters closer to the tower, the angle of elevation is 85 degree. Determine the acute angle from the horizontal at which the tower is leaning?
Answers
Answered by
Reiny
First observation point A
second observation point B
top of tower P
bottom of tower Q
In triangle ABP,
angle A = 70 --- given
angle PBA = 95 --- exterior angle to 85°
angle APB = 15°
AB = 55m --- given
by the sine law
BP/sin70 = 55/sin15
BP = 55sin70/sin15
in triangle BPQ
PQ=200
angle PBQ = 85
BP = 55sin70/sin15
sinQ/BP= sin85/200
sinQ = BPsin85/200
= (55sin70/sin15)(sin85)/200 = .9946..
angle Q = 84.0658°
or appr 84°
second observation point B
top of tower P
bottom of tower Q
In triangle ABP,
angle A = 70 --- given
angle PBA = 95 --- exterior angle to 85°
angle APB = 15°
AB = 55m --- given
by the sine law
BP/sin70 = 55/sin15
BP = 55sin70/sin15
in triangle BPQ
PQ=200
angle PBQ = 85
BP = 55sin70/sin15
sinQ/BP= sin85/200
sinQ = BPsin85/200
= (55sin70/sin15)(sin85)/200 = .9946..
angle Q = 84.0658°
or appr 84°
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