Question

From the top of a building 60m high, the angles of depression of the top and bottom of a vertical

lamp post are observed to be 30° and 60°respectively. Find [i] horizontal distance between the

building and the lamp post [ii] 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 the lamp post =20 √3 cm

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

Hope it helps...
Thank You !

Answer 2

Answer:

Step-by-step explanation:

the answer is 40m