Question : The angle of elevation of an aeroplane as observed from a point 30 metres above the transparent water surface of a lake is 30° and the angle of the depression of the image of the aeroplane in the water of the lake is 60°. The height of the aeroplane from the water surface of the lake is:
Option 1: 60 metres
Option 2: 45 metres
Option 3: 50 metres
Option 4: 75 metres
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Correct Answer: 60 metres
Solution : AB = transparent water surface then according to question, $\triangle$CPM = 30°; $\angle$C'PM = 60° CM = $h$ CB = Height of the aeroplane from the water surface of the lake = $h + 30$ $\therefore$ C'B = $h + 30$ In $\triangle$CMP, tan 30° = $\frac{CM}{PM}$ $⇒\frac{1}{\sqrt{3}} = \frac{h}{PM}$ $\therefore$ PM = ${\sqrt{3}}$h------------(1) In $\triangle$PMC', tan 60° = $\frac{C'M}{PM}$ $⇒{\sqrt{3}} = \frac{h+30+30}{PM}$ $\therefore$ PM = $\frac{h+60}{\sqrt{3}}$-----------(2) From equation 1 and 2, we get, $3h = h+60$ $⇒ h = 30 $ $\therefore$ CB = BM + CM = 30 + 30 = 60 metres Hence, the correct answer is 60 metres.
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