Choose the correct answer of the following question:
From a point on the ground, 30 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The height of the
tower is
(a) $30 \mathrm{~m}$
(b) $10 \sqrt{3} \mathrm{~m}$
(c) $10 \mathrm{~m}$
(d) $30 \sqrt{3} \mathrm{~m}$
Let AB be the tower and point C be the point of observation on the ground.
We have,
$\mathrm{BC}=30 \mathrm{~m}$ and $\angle \mathrm{ACB}=30^{\circ}$
In $\Delta \mathrm{ABC}$,
$\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{30}$
$\Rightarrow \mathrm{AB}=\frac{30}{\sqrt{3}}$
$\Rightarrow \mathrm{AB}=\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow \mathrm{AB}=\frac{30 \sqrt{3}}{3}$
$\therefore \mathrm{AB}=10 \sqrt{3} \mathrm{~m}$
Hence, the correct answer is option (b).
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