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?
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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