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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 $45^\circ$ and angle of depression of its foot is $30^\circ$.
(i) Write a relation between $d$ (the height of window) and $y$.
(ii) Determine the value of $h$.
(iii) (a) Determine height of the water tank.
OR
(iii) (b) Find the value of $x$ and height of the window above ground level.
From a window (W) of the building, the angle of elevation of top of the tank is $45^\circ$ and angle of depression of its foot is $30^\circ$.
(i) Write a relation between $d$ (the height of window) and $y$.
(ii) Determine the value of $h$.
(iii) (a) Determine height of the water tank.
OR
(iii) (b) Find the value of $x$ and height of the window above ground level.
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Sol. (i) $\sin 30^\circ = \frac{d}{y} \Rightarrow 2d = y$ (I Mark)
(ii) $\tan 45^\circ = 1 = \frac{h}{WX} = \frac{h}{54} \Rightarrow h = 54$ m (I Mark)
(iii) (a) $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{d}{54} \Rightarrow d = 18\sqrt{3}$ m (I Mark)
Height of the tank = $h + d = (54 + 18\sqrt{3})$ m (II Mark)
OR
(iii) (b) $\angle WAC = 30^\circ, \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{WC}{54} \Rightarrow WC = 18\sqrt{3}$ m (I Mark)
$\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{h}{x} \Rightarrow x = h\sqrt{2} \Rightarrow x = 54\sqrt{2}$ m (II Mark)
(ii) $\tan 45^\circ = 1 = \frac{h}{WX} = \frac{h}{54} \Rightarrow h = 54$ m (I Mark)
(iii) (a) $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{d}{54} \Rightarrow d = 18\sqrt{3}$ m (I Mark)
Height of the tank = $h + d = (54 + 18\sqrt{3})$ m (II Mark)
OR
(iii) (b) $\angle WAC = 30^\circ, \tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{WC}{54} \Rightarrow WC = 18\sqrt{3}$ m (I Mark)
$\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{h}{x} \Rightarrow x = h\sqrt{2} \Rightarrow x = 54\sqrt{2}$ m (II Mark)