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°
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?
1 answer