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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