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

Question

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 $30°$ at a particular instant.

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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 $\sqrt{3} \approx 1.73$)
(iii) (a) Find the time (in seconds) in which the angle of depression changes from $30°$ to $45°$.

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 $60°$ from the drone?

Accepted Answer

Step 1: Right triangle with Q at top, P at bottom-left, R at bottom-right. QP = 100 m (vertical), PR = d (horizontal), angle of depression from Q = 30°, angle QRP = 30°
Step 2: (ii) In $	riangle PQR$, $\frac{100}{d} = 	an 30° = \frac{1}{\sqrt{3}}$
$\Rightarrow d = 100\sqrt{3} = 173$ m
Step 3: Triangle diagram with Q at top, S extending right from Q, P at bottom-left, M and R on the base line. Angles 30° and 45° at Q, angles 45° and 30° at M and R. PM = 173-x, MR = x
Step 4: (iii)(a) In $	riangle PQM$, $\frac{100}{173 - x} = 	an 45° = 1$
$\Rightarrow x = 73$ m
Time taken $= \frac{73 	imes 18}{60 	imes 5} = \frac{219}{50}$ or $4.4$ seconds (approx.)

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Solution