test question

In triangle ABC,

tan 45^o = AB/ BC = 50/ BC

1 = 50/ BC

BC = 50

Let the height of the building be h.

Now, In triangle ADE

tan 30^o = AE/ DE             [AE = 50 - h, DE = 50m]

1/3 = (50 - h)/ 50

50/ 3 = 50 - h

h = 50 - 50/ 3

So, the height of building is 50 - 50/ 3

Horizontal distance of point of observation from top of the building = AD

cos 30^o = DE/ AD

3 / 2 = 50 / AD

Therefore, AD = 100/3

Try on similar  lines to get  distance of top of cliff from the bottom of building.