The angle of elevation of an airborne helicopter from a point A on the ground is 45° . After a flight of 15 seconds,…
CBSE Class 10 Maths PYQ · Applications of Trig · Speed Distance · 5 Marks · March 2025 · Standard
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705 Marks · March 2025 · Standard
The angle of elevation of an airborne helicopter from a point $A$ on the ground is $45^\circ$. After a flight of $15$ seconds, the angle of elevation of the helicopter changes to $30^\circ$. If the helicopter is flying at a constant height of $2000$ m, find the speed of the helicopter. (Take $\sqrt{3} = 1.732$)
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Correct figure ($1$ mark) In right $\triangle APB$ $\frac{2000}{AP} = \tan 45^\circ = 1$ $\Rightarrow AP = 2000$ m ($1/2$ mark) In right $\triangle ACD$ $\frac{2000}{AC} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow AC = 2000 \sqrt{3}$ m ($1/2$ mark) $BD = PC = AC - AP = 2000 \sqrt{3} - 2000$ ($1$ mark) $= 2000 (1.732 - 1)$ $= 1464$ m ($1$ mark) Time taken from $B$ to $D = 15$ sec Speed $= \frac{1464}{15} = 97.6$ m/s ($1$ mark)
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