A 5-metre plank rests on a wall

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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