From a point on the ground, …

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Gaurav Seth 5 years, 3 months ago
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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