Question : A pilot in an aeroplane at an altitude of 200 metres observes two points lying on either side of a river. If the angles of depression of the two points be $45^{\circ}$ and $60^{\circ}$, then the width of the river is:
Option 1: $(200+\frac{200}{\sqrt{3}})$ metres
Option 2: $(200-\frac{200}{{\sqrt3}})$ metres
Option 3: $400 {\sqrt3}$ metres
Option 4: $(\frac{400}{{\sqrt3}})$ metres
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Correct Answer: $(200+\frac{200}{\sqrt{3}})$ metres
Solution :
Let BD be $x$ and CD be $y$.
In right $\triangle$ABD,
$\tan 45^{\circ}$ = $\frac{AD}{BD}$
$⇒1 = \frac{200}{x}$
$⇒x= 200$
Now, in right $\triangle$ACD,
$\tan 60^{\circ}= \frac{AD}{CD}$
$⇒\sqrt3= \frac{200}{y}$
$⇒y= \frac{200}{\sqrt3}$
$\therefore$ The width of the river $=x+y = 200+\frac{200}{\sqrt3}$ metres
Hence, the correct answer is $200+\frac{200}{\sqrt3}$ metres.
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