From the top of building 60m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° resp. Find 1) the horizontal distance between the building and the lamp post. 2) the ht of lamp post ( use √3 = 1.732)
Answers
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Let AB be the building
And CD be the lamp post. While DE is the horizontal line parallel to the ground from the top of the lamp post to the building.
So in Triangle ABC,
AB=60m
θ=60°
tan θ = perpendicular /base
tan 60°= AB / BC
√3=60 / BC
BC = 60 / √3
On rationalising denominator, we get
BC=60 * √3 /3
=[20 * √3]m
Distance between building and lamp post =20 √3 cm
=20*1.732(√3=1.732)
=34.64m(Ans)
EBCD is a rectangle , hence BC =ED
In Triangle AED
θ=30°
tan 30° = AE / ED
1 /√3 = AE / 20√3AE * √3 = 20√3AE = 20mAB = AE + EBEB =60 - 20=40m
Since EB = CD (EBCD being a rectangle)CD or Height of lamp post = 40m (Ans)
