Question

From the top of the building 60m high the angle of depression of the top and bottom of a vertical lamp post are observed to be 30 degree and 60 degree respectively find a)the distance between the building and the lamp post b)the height of the lamp post

Answer 1

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)      

Answer 2

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)      

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