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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?

  • trig -

    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.

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