Question : From the top of an upright pole $24 \sqrt{3}$ feet high, the angle of elevation of the top of an upright tower was $60^{\circ}$. If the foot of the pole was 60 feet away from the foot of the tower, how tall (in feet) was the tower?
Option 1: $84 \sqrt{3}$
Option 2: $36\sqrt{3}$
Option 3: $44\sqrt{3}$
Option 4: $60\sqrt{3}$
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Correct Answer: $84 \sqrt{3}$
Solution :
Here, AB = $24\sqrt3$ feet = PQ
Distance, AP = 60 feet
Here, AP = BQ = 60 feet
In $\triangle$BQR,
$\tan 60° = \frac{RQ}{BQ}$
⇒ $\sqrt3=\frac{RQ}{60}$
$\therefore RQ = 60\sqrt3$
Height of the tower = RP = RQ + PQ = $60\sqrt3+24\sqrt3=84\sqrt3$ feet
Hence, the correct answer is $84\sqrt3$ feet.
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