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A 5-metre plank rests on a wall

2-m high,so that 1.5 m of the plank projects beyond the wall.
A. What angle does the plank make with the wall?
B. How high is the end of the plank above the ground?

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1 answer
  1. Draw horizontal line.

    Mark the starting point with A

    In point A draw a straight line with some angle.

    Select a point on the horizontal line and mark it with B.

    Draw a vertical line in point B (this vertical line is the wall).

    Mark the point where the vertical line touches a straight line with C.

    θ is angle which the plank make with the wall.

    Extend the AC line and mark one point with D.

    From point D, draw a vertical line that touches the horizontal line
    and mark the point of contact with E.

    2 m high, so that 1.5 m of the plank projects beyond the wall means:

    BC = 2 m

    AC = 5 - 1.5 = 3.5 m

    cos θ = BC / AC = 2 / 3.5 = 0.571428571

    θ = cos⁻¹ ( 0.571428571 ) = 55.1501° = 55° 9'

    From the similarity of triangles:

    BC / AC = DE / AD

    Since BC = 2 m , AC = 3.5 m , AD = 5 m:

    2 / 3.5 = DE / 5

    Cross multiply.

    10 = 3.5 ∙ DE

    DE = 10 / 3.5 = 2.8571428571

    DE ≈ 2.86 m

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    Remark:

    cos⁻¹ sometimes written as arccos so,

    θ = cos⁻¹ ( 0.571428571 ) = 55.1501° = 55° 9'

    is the same as

    θ = arccos ( 0.571428571 ) = 55.1501° = 55° 9'

    ≈ means approximately equal
    ______________________________________

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