trigonometry(height and distance)

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The shadow of a tower when the angle of elevation of the sun is 45 degree is found to be 5m longer when it is 60 degree. Find the height of the tower.

  • trigonometry(height and distance) -

    How about starting with a diagram.
    Label the tower as PQ, with Q on the ground.
    Label the point with angle 45° as A, and the point with the 60° angle as B.
    You now have two right-angled triangles, PAQ and PBQ

    let BQ be x
    then tan60 = PQ/x ---> PQ = xtan60 = √3 x
    and tan 45 = PQ/(x+5) ---> PQ = (x+5)tan45 = (x+5)(1)

    so √3 x =x+5
    √3 x - x = 5
    x(√3 - 1) = 5
    x = 5/(√3-1)
    then PQ = √3(5/(√3-1)) = appr 11.83 m

    or

    look at triangle PAB, angle PBA = 120° making angle APB = 15°
    by the sine law:
    PB/sin45 = 5/sin15
    PB = 5sin45/sin15

    in triangle PBQ,
    sin60 = PQ/PB
    PQ = PBsin60 = (5sin45/sin15)(√3/2) = appr 11.83

    In my experience, most students find the second method easier to follow

  • trigonometry(height and distance) -

    Ans is correct can you show me fig.

  • trigonometry(height and distance) -

    We can't put diagrams on here, I think you should be able to follow my directions. I described the diagram.

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