two poles of equal heights are on 2sides of highway 30m abide. the angle of elevation of the top of 2 poles from a point between them are such that their tangent ratios are 1 and 1/2 find their height?
Answers
Let AB and CD be the two poles of equal height standing on the two sides of the road of width 80 m.
AC = 80 m Let AP = x then PC = 80 – x P is a point on the road from which the angle of elevation of the top of tower AD is 60°, also, the angle of depression of the point P from the point B is 30°.
∠BPA = 60° and ∠DPC = 30° Draw a parallel line from DS to CA. ∠SDP = ∠DPC = 30°.
To Find : The height of the poles and the distance of the point P from both the poles.
From right
△PAB: tan 60° = AB/AP √3 = h/x or h = x√3 …(1)
From right △PCD: tan 30° = AB/AP 1/√3 = h/(80-x) or h = (80-x)/√3 …(2) Form (1) and (2),
we have (80-x)/√3 = x√3 80 – x = 3x x = 20
Form (1): h = (20)√3
Therefore, Height of each pole = (20)√3 m Distance of pole AB from point P = 20 m Distance of pole CD from point P = 80 – 20 = 60.
Answer:
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