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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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?
1 answer