You need to clarify the diagram. Label the airplane A, the tower T and the post P. The plane is at altitude h, with AP=t and AT=p. Point Q is directly below the plane. The distance PQ=q and thus TQ = 2-q.
If the plane is between the tower and the post, then
h/q = tanα
h/(2-q) = tanθ
Eliminate h and plug in the angles, and to get the desired distance t,
q tan 81° = (2-q)tan42°
6.31q = 0.90(2-q)
q = 0.249
so, h = 0.249*6.31 = 1.571
t^2 = .249^2 + 1.571^2
t = 1.59 miles
If the plane is not between the two measurement locations, then you need to set up different triangles.
The angles of elevation theta and alpha to an airplane from the airport control tower and from an observation post 2 miles away are being continuously monitored. If theta is 42 degrees when alpha is 81 degrees, how far is the plane from the observation post?
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