Question

plZz solve this...............​

Answer 1

Answer:

hope it helps you see the attachment for further information

Answer 2

Answer:

[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.

HOPE THIS WILL HELP YOU...