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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 4 · Set 1 · Q38
A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance…
A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of $\displaystyle 60$ km/h. The drone stopped at a point Q, $\displaystyle 100$ m vertically above the point $\displaystyle P$. The angle of depression of the ambulance was found to be $\displaystyle 30$° at a particular instant. Based on above information, answer the following questions :
(i)
Represent the above situation with the help of a diagram.
(ii)
Find the distance between the ambulance and the site of accident (P) at the particular instant. (Use $\displaystyle \sqrt{3}=1.73$ )
(iii)
(a)
Find the time (in seconds) in which the angle of depression changes from $\displaystyle 30$° to $\displaystyle 45$°.
OR
(b)
How long (in seconds) will the ambulance take to reach point P from a point T on the highway such that angle of depression of the ambulance at T is $\displaystyle 60^{\circ}$ from the drone ?
Marking-scheme solution
(i) 
(ii) In \(\displaystyle \Delta \mathrm{PQR}, \frac{100}{\mathrm{~d}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
\[\Rightarrow d=100 \sqrt{3}=173 m
\]
(iii)
For correct figure
In \(\displaystyle \Delta \mathrm{PQM}, \frac{100}{173-x}=\tan 45^{\circ}=1\)
\(\displaystyle \Rightarrow x=73 \mathrm{~m}\)
Time taken \(\displaystyle =\frac{73 \times 18}{60 \times 5}=\frac{219}{50}\) or $\displaystyle 4.4$ seconds (approx.)
For correct figure
In \(\displaystyle \Delta \mathrm{PQT}, \frac{100}{y}=\tan 60^{\circ}=\sqrt{3}\)
\(\displaystyle \Rightarrow \mathrm{y}=\frac{100}{\sqrt{3}}=\frac{100 \sqrt{3}}{3}\) or \(\displaystyle \frac{173}{3} \mathrm{~m}\)
Time taken \(\displaystyle =\frac{100 \sqrt{3} \times 18}{3 \times 60 \times 5}=2 \sqrt{3}\) or $\displaystyle 3.5$ seconds (approx.)
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