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: $\frac{40 \sqrt{3}}{3}$
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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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