Question : From an aeroplane just over of river, the angle of depression of two palm trees on the opposite bank of the river is found to be $60^{\circ}$ and $30^{\circ}$, respectively. If the breadth of the river is 400 metres, then the height of the aeroplane above the river at that instant is: $(\sqrt{3} = 1.732)$
Option 1: 173.2 metres
Option 2: 346.4 metres
Option 3: 519 .6 metres
Option 4: 692.8 metres
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Correct Answer: 173.2 metres
Solution :
Let,
$BC$ = River = $400$ m
$AD$ = Height of plane = $H$ m
$BD=x$ m
$CD=(400-x)$ m
From $\Delta ABD$,
$\tan 60^{\circ} = \frac{AD}{BD}$
$⇒\sqrt3 = \frac{H}{x}$
$⇒H = \sqrt3x$
$⇒x=\frac{H}{\sqrt3}$--------------(i)
From $\Delta ADC$,
$\tan 30^{\circ} = \frac{AD}{CD}$
$⇒\frac{1}{\sqrt3} = \frac{H}{400-x}$
$⇒\sqrt3H = 400-x$
Putting the value of $x$ from equation (i), we get,
$⇒\sqrt3H = 400-\frac{H}{\sqrt3}$
$⇒4H=400\sqrt3$
$⇒H= 100\sqrt3 $
$\therefore H=173.2$ [As $\sqrt3=1.732$]
Hence, the correct answer is 173.2 metres.
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