The angle of elevation of an aeroplane from a point on the ground is 45° . After a flight of 15 seconds, the elevation…
CBSE Class 10 Maths PYQ · Applications of Trig · Speed Distance · 5 Marks · July 2023 · Standard
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665 Marks · July 2023 · Standard
The angle of elevation of an aeroplane from a point on the ground is $45^\circ$. After a flight of $15$ seconds, the elevation changes to $30^\circ$. If the aeroplane is flying at a constant height of $3000$ meters, find the speed of the aeroplane in km/h. [Take $\sqrt{3} = 1.732$]
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Let the height of the aeroplane be $h = 3000$ m. Let the initial position of the aeroplane be A and final position be B. Let the point on the ground be P. In $\triangle APM$, $\tan 45^\circ = \frac{AM}{PM} = \frac{3000}{x}$ $1 = \frac{3000}{x} \Rightarrow x = 3000$ m. In $\triangle BPN$, $\tan 30^\circ = \frac{BN}{PN} = \frac{3000}{x+y}$ $\frac{1}{\sqrt{3}} = \frac{3000}{3000+y} \Rightarrow 3000+y = 3000\sqrt{3}$ $y = 3000\sqrt{3} - 3000 = 3000(\sqrt{3}-1)$ $y = 3000(1.732-1) = 3000(0.732) = 2196$ m. Distance covered by aeroplane in $15$ seconds is $y = 2196$ m. Speed of aeroplane $= \frac{\text{Distance}}{\text{Time}} = \frac{2196 \text{ m}}{15 \text{ s}}$ Speed $= \frac{2196}{15}$ m/s $= 146.4$ m/s. To convert to km/h: $146.4 \times \frac{3600}{1000} = 146.4 \times 3.6 = 527.04$ km/h.
