Question

Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is 54 metres away from the water tank.  From a window (W) of the building, the angle of elevation of top of the tank is 45° and angle of depression of its foot is 30°.

(i) Write a relation between d (the height of window) and y.   [1]

(ii) Determine the value of h.   [1]

(iii) (a) Determine height of the water tank.   [2]

OR

(iii) (b) Find the value of x and height of the window above ground level.   [2]

Answer 1

(i) In ΔWAC

`sin 30^circ = P/H`   ...[∠XWA = ∠WAC = 30°] [Alternate interior angles]

⇒ `1/2 = d/y`   ...[WC = XA = d]

⇒ `d = y/2`

(ii) In ΔBWX

`tan 45^circ = h/54`

⇒ h = 54 m

(iii) (a) Height of water tank = h + d = 54 + d

Now, In ΔWCA

`tan 30^circ = (WC)/(AC) = d/54`

`1/sqrt(3) = d/54`

`d = 54/sqrt(3) m`

= `(54sqrt(3))/3`

= `18sqrt(3)  m`

Height of water tank = `54 + 18sqrt(3)`

= `18(3 + sqrt(3)) m`

= `18sqrt(3)(sqrt(3) + 1) m`

OR

(iii) (b) `cos 45^circ = (XW)/(BW) = B/H`

⇒ `1/sqrt(2) = 54/x`

⇒ `x = 54sqrt(2)  m`

Height of window = d

In ΔWCA

`tan 30^circ = (WC)/(AC) = d/54`

⇒ `1/sqrt(3) = d/54`

⇒ `d = 54/sqrt(3) xx sqrt(3)/sqrt(3)`

= `(54sqrt(3))/3`

= `18sqrt(3)  m`