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March 2, 2015

March 2, 2015

Posted by **Greg** on Monday, February 14, 2011 at 5:50pm.

- trig -
**MathMate**, Monday, February 14, 2011 at 7:11pmLet 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.

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