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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Question : The angle of elevation of an aeroplane from a point on the ground is 60°. After flying for 30 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a height of 4500 metres, then what is the speed (in m/s) of an aeroplane?
Option 1: $50\sqrt3$
Option 2: $100\sqrt3$
Option 3: $200\sqrt3$
Option 4: $300\sqrt3$
Question : The angles of elevation of an aeroplane flying vertically above the ground, as observed from the two consecutive stones,1 km apart, are 45° and 60°. The height of the aeroplane above the ground in km is:
Option 1: $(\sqrt{3}+1)$
Option 2: $(\sqrt{3}+3)$
Option 3: $\frac{1}{2}(\sqrt{3}+1)$
Option 4: $\frac{1}{2}(\sqrt{3}+3)$
Question : An aeroplane flying horizontally at a height of 3 km above the ground is observed at a certain point on earth to subtend an angle of $60^\circ$. After 15 seconds of flight, its angle of elevation is changed to $30^\circ$. The speed of the aeroplane (Take $\sqrt{3}=1.732$) is:
Option 1: 230.63 m/s
Option 2: 230.93 m/s
Option 3: 235.85 m/s
Option 4: 236.25 m/s
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
Question : The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 metres longer than when it was 60°. The height of the tower is:
Option 1: $5\sqrt{3}-1$ metres
Option 2: $5(3+\sqrt{3})$ metres
Option 3: $10(\sqrt{3}-1)$ metres
Option 4: $10(\sqrt{3}+1)$ metres
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