Question : From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60°. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
Option 1: $40 \sqrt{3}$
Option 2: $\frac{151 \sqrt{3}}{6}$
Option 3: $\frac{77}{4} \sqrt{3}$
Option 4: $\frac{40 \sqrt{3}}{3}$
Correct Answer: $\frac{40 \sqrt{3}}{3}$
Solution :
We have to find the value of $x$.
In $\triangle ABC,$
$\tan60° = \frac{AC}{BC}$
We know, $CE=BD$
⇒ $AC=AE-CE$
⇒ $AC=57.75-17.75$
⇒ $AC=40$
In $\triangle ABC,$
$\tan60° = \frac{40}{x}$
⇒ $\sqrt3=\frac{40}{x}$
⇒ $x=\frac{40}{\sqrt3}$
⇒ $x=\frac{40\sqrt3}{3}$ m
Hence, the correct answer is $\frac{40\sqrt3}{3}$.
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