Question : The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 metres, then the speed of the aeroplane in km/h is:
Option 1: $600$
Option 2: $600(\sqrt{3}+1)$
Option 3: $600\sqrt{3}$
Option 4: $600(\sqrt{3}–1)$
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Correct Answer: $600(\sqrt{3}–1)$
Solution :
Let the vertical height be $h$ and horizontal distance be $B_{1}$ when the elevation is 45° and $B_{2}$ when the elevation is 30°.
Using trigonometry ratio, when elevation is 45°.
$\tan\theta =\frac{\text{Height}}{\text{Distance}}$
⇒ $\tan45° = \frac{h}{B_{1}}$
⇒ $1 = \frac{h}{B_{1}}$
⇒ $B_{1} = h$
Using trigonometry ratio, when elevation is 30°.
$\tan30° = \frac{h}{B_{2}}$
⇒ $\frac{1}{\sqrt{3}} = \frac{h}{B_{2}}$
⇒ $B_{2} = \sqrt{3}h$
The speed of the aeroplane = $\frac{B_{2}–B_{1}}{\text{time}}$
Also, time is given as 15 sec and height is 2500 metres.
So, speed = $ \frac{B_{2}–B_{1}}{\text{time}} = \frac{\sqrt{3}h–h}{\frac{15}{60×60}}$ $\frac{(\sqrt{3}–1)h}{\frac{15}{60×60}} = \frac{(\sqrt{3}–1)\frac{2500}{1000}}{\frac{15}{60×60}}$ $=600(\sqrt{3}–1)$ km/h
Hence, the correct answer is $600(\sqrt{3}–1)$ km/h.
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