At 2 points 95 ft apart on a horizontal line perpendicular to the front of a building, the angles of elevation of the top of the building are 25 degrees and 16 degrees. How tall is the building?

Answer 1

Let the two points be A, and B 95 feet apart, with B closer to the building.

Let the line join the building's base at P, and the top of the building be Q.

Let the horizontal distance BP=x, and
the vertical distance
= height of building
= h

So
BPQ is a right-triangle, with QBP=25°, and ∠BPQ 90°.

So the
h = xtan(25°) .... (1)

But AP=AB+BP=95'+x
so
(95'+x)tan(16°) = h ...(2)

Solve for x and h to get:
x=151.7'
h=70.7'

Check my work.