The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, then find the height of the building.
Let AB be the building and PQ be the tower.
We have,
$\mathrm{PQ}=60 \mathrm{~m}, \angle \mathrm{APB}=30^{\circ}, \angle \mathrm{PAQ}=60^{\circ}$
In $\Delta \mathrm{APQ}$,
$\tan 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AP}}$
$\Rightarrow \sqrt{3}=\frac{60}{\mathrm{AP}}$
$\Rightarrow \mathrm{AP}=\frac{60}{\sqrt{3}}$
$\Rightarrow \mathrm{AP}=\frac{60 \sqrt{3}}{3}$
$\Rightarrow \mathrm{AP}=20 \sqrt{3} \mathrm{~m}$
Now, in $\triangle \mathrm{ABP}$,
$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{20 \sqrt{3}}$
$\Rightarrow \mathrm{AB}=\frac{20 \sqrt{3}}{\sqrt{3}}$
$\therefore \mathrm{AB}=20 \mathrm{~m}$
So, the height of the building is 20 m.
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