A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there…
CBSE Class 10 Maths PYQ · Applications of Trig · Speed Distance · 4 Marks · March 2025 · Standard
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654 Marks · March 2025 · Standard
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 $60$ km/h. The drone stopped at a point Q, $100$ m vertically above the point P. The angle of depression of the ambulance was found to be $30^\circ$ 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 $\sqrt{3}= 1.73$) (iii) (a) Find the time (in seconds) in which the angle of depression changes from $30^\circ$ to $45^\circ$. OR (iii) (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 $60^\circ$ from the drone ?
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(i) For correct figure (ii) In $\triangle PQR$, $\frac{100}{d} = \tan 30^\circ = \frac{1}{\sqrt{3}}$ $\Rightarrow d = 100\sqrt{3} = 173$ m (iii) (a) For correct figure In $\triangle PQM$, $\frac{100}{173-x} = \tan 45^\circ = 1$ $\Rightarrow x = 73$ m Time taken = $\frac{73\times 18}{60\times5} = \frac{219}{50}$ or $4.4$ seconds (approx.) OR (iii) (b) For correct figure In $\triangle PQT$, $\frac{100}{y} = \tan 60^\circ = \sqrt{3}$ $\Rightarrow y = \frac{100}{\sqrt{3}} = \frac{100\sqrt{3}}{3}$ or $173/3$ m Time taken = $\frac{100\sqrt{3} \times 18}{3 \times 60 \times5} = 2\sqrt{3}$ or $3.5$ seconds (approx.)
