Question : Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45°, respectively. If the height of the tower is 50 metres, the distance between the two men is: (take $\sqrt{3}=1.732$)
Option 1: $136.6$ metres
Option 2: $50\sqrt{3}$ metres
Option 3: $100\sqrt{3}$ metres
Option 4: $135.5$ metres
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Correct Answer: $136.6$ metres
Solution :
Let the positions of the two men be A and B.
Let PQ represent the tower.
$\angle$QAP = 30° and $\angle$QBP = 45°
QP = 50 m
We get two right-angled triangles △QPB and △QPA
Now, From the △QPA:
tan 30° = $\frac{PQ}{AP}$
⇒$\frac{1}{\sqrt{3}}=\frac{50}{AP}$
⇒ AP = $50\sqrt{3}=50×1.732=86.6$ metres
From the △QPB:
⇒ $\tan 45° = \frac{PQ}{PB}$
⇒ 1 = $\frac{50}{PB}$
⇒ PB = 50 metres
Now, AB = AP + PB
= 86.6 + 50
= 136.6 metres
Hence, the correct answer is 136.6 metres.
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