trig
posted by Greg .
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?

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 righttriangle, 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.
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