Precalculus With Limits, Chapter 6, 6.1, Section 6.1, Problem 54

The angles of elevation to an airplane from two points A and B on level ground are 55 degrees and 72 degrees respectively. The points A and B are 2.2 miles apart, and the airplane is east of both points in the same vertical plane.
Let the altitude of the plane be H. Let the point at which the plane is be represented by P, and the point where the vertical touches the ground be Q.
The points A, Q and P form a right triangle with right angle /_ AQP. And the points B, Q and P form a right triangle with right angle at /_ BQP.
PQ = BQ*tan 55 and PQ = AQ*tan 72.
BQ = AQ + 2.2
Substituting this in the formulas derived earlier.
(AQ + 2.2)*tan 55 = AQ*tan 72
AQ*(tan 72 - tan 55) = 2.2
AQ = 2.2/(tan 72 - tan 55) ~~ 1.33
PQ = 1.33*tan 55 = 1.9
The altitude of the plane is approximately 1.9 miles.