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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