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The angles of depression of the top and the foot of a $9 \text{ m}$ tall building from the top of a multi-storeyed building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $\sqrt{3} = 1.73$)
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Let CD be the multi storeyed building of height $h$ metres and AB be the $9 \text{ m}$ tall building.
In $\triangle DEB, \tan 30^\circ = \frac{h - 9}{x} \Rightarrow x = (h - 9)\sqrt{3} \dots(i)$ ($1 + \frac{1}{2}$ marks)
In $\triangle DCA, \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \dots(ii)$ ($1 + \frac{1}{2}$ marks)
Solving (i) & (ii) we get, $x = 7.79, h = 13.5$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$\therefore \text{The height is } 13.5 \text{ m and the distance is } 7.79 \text{ m}$
In $\triangle DEB, \tan 30^\circ = \frac{h - 9}{x} \Rightarrow x = (h - 9)\sqrt{3} \dots(i)$ ($1 + \frac{1}{2}$ marks)
In $\triangle DCA, \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \dots(ii)$ ($1 + \frac{1}{2}$ marks)
Solving (i) & (ii) we get, $x = 7.79, h = 13.5$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$\therefore \text{The height is } 13.5 \text{ m and the distance is } 7.79 \text{ m}$