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Mathematics · 2025 · 5 marks
CBSE 2025 · Region 2 · Set 2 · Q35
(a) The angle of elevation of an airborne helicopter from a point A on the ground is 45°. After a flight of 15 seconds, the angle of elevation of the…
(a)
The angle of elevation of an airborne helicopter from a point A on the ground is $\displaystyle 45$°. After a flight of $\displaystyle 15$ seconds, the angle of elevation of the helicopter changes to $\displaystyle 30$°. If the helicopter is flying at a constant height of $\displaystyle 2000$ m, find the speed of the helicopter. (Take $\displaystyle \sqrt{3}=1 \cdot 732$ )
OR
(b)
A girl $\displaystyle 1.5$ m tall is standing at some distance from a $\displaystyle 30$ m high tower. The angle of elevation from her eye to the top of the tower increases from $\displaystyle 30$° to $\displaystyle 60$° as she walks towards the tower. Find the distance she walked towards the tower.
Marking-scheme solution
In right \(\displaystyle \Delta \mathrm{APB}\)
\(\displaystyle \Rightarrow \mathrm{AP}=2000 \mathrm{~m}\)
In right \(\displaystyle \Delta \mathrm{ACD}\)
\(\displaystyle \frac{2000}{\mathrm{AC}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
\(\displaystyle \Rightarrow \mathrm{AC}=2000 \sqrt{3} \mathrm{~m}\)
\[\begin{aligned}
\mathrm{BD}=\mathrm{PC}=\mathrm{AC}-\mathrm{AP} & =2000 \sqrt{3}-2000 \\
& =2000(1.732-1) \\
& =1464 \mathrm{~m}
\end{aligned}
\]
Time taken from B to \(\displaystyle \mathrm{D}=15 \mathrm{sec}\)
\[\text { Speed }=\frac{1464}{15}=97.6 \mathrm{~m} / \mathrm{s}
\]
\(\displaystyle \mathrm{AF}=30-1 \cdot 5=28 \cdot 5 \mathrm{~m}\)
In right \(\displaystyle \Delta\) AFE
\(\displaystyle \frac{28.5}{\mathrm{EF}}=\tan 60^{\circ}=\sqrt{3}\)
\(\displaystyle \Rightarrow \mathrm{EF}=\frac{28.5}{\sqrt{3}} \mathrm{~m}\)
In right \(\displaystyle \Delta\) ADF
\(\displaystyle \frac{28.5}{\mathrm{DF}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
\(\displaystyle \Rightarrow \mathrm{DF}=28.5 \sqrt{3} \mathrm{~m}\)
Distance travelled by the girl towards the tower, \(\displaystyle \mathrm{DE}=\mathrm{DF}-\mathrm{EF}\)
\[\begin{aligned}
& =28.5 \sqrt{3}-\frac{28.5}{\sqrt{3}} \\
& =\frac{57}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\
& =19 \sqrt{3} \mathrm{~m} \text { or } 32.91 \mathrm{~m} \text { approx. }
\end{aligned}
\]
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