Question

From a point on the ground the angle of elevation of the bottom and top of the water tank at the top of 20 meters high tower are 45 degrees and 60 degrees. Draw the diagram

Answer 1

According to question,

Height of building  L = 20 m

Angle from top of building   theta = 45 deg
Angle from top of tank     A = 60 deg

Let the height of tank be x

Thus

tan theta = L/ B
tan 45 = 20 / B
Thus 

B = 20 m
Now

tan A = x + L / B
x + L = tan 60 * 20
x  =  20 ( root 3 - 1)

Answer 2

[FIGURE IS IN THE ATTACHMENT]
Let AB and BC be the tower and water tank and D be the point of observation.
Given:
BC = 20m
∠BDC = 45° and ∠ADC = 60°
Let AB = h m
In ΔBDC
tan 45° = P/B = BC/DC
1 = 20/DC [ tan 45°=1]
DC= 20 m
Now in ΔADC
tan 60°= P/B = AC /DC = (AB + BC)/DC
√3 = (h+20)/20
20√3 = h+20
20√3 - 20 = h
20(√3-1)= h
h= 20(√3-1) m

h = 20 (1.73 -1)
h = 20 × .73 = 14.6 m

Hence, the height of the tower is 20(√3-1) or 14.6 m.

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