Question

A drone was used to facilitate the movement of an ambulance on a straight highway to a point P on the ground where there was an accident. The ambulance was travelling at a 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.

Based on the above information, answer the following questions:

1. Represent the above situation with the help of a diagram. (1)
2. Find the distance between the ambulance and the site of the accident (P) at the particular instant. (Use `sqrt3` = 1.73) (1)
3. 1. Find the time (in seconds) in which the angle of depression changes from 30° to 45°. (2)
      OR
   2. How long (in seconds) will the ambulance take to reach point P from a point T on the highway, such that the angle of depression of the ambulance at T is 60° from the drone? (2)

Answer 1

i.

ii. In APQR,

tan 30 = `(PQ)/(PR)`

`1/sqrt3 = 100/(PR)`

PR = `100sqrt3`

PR = 100 × 1.73

= 173 m

iii. a.

Speed = 60 km/h

= `(60 xx1000)/(60 xx 60)`

= `100/6` m/sec

In ΔPQR

tan 45 = `(PQ)/(PR')`

1 = `100/(PR')`

PR' = 100 m

x = 100 m

tan 30 = `(PQ)/(PR)`

`1/sqrt3 = 100/(x + y)`

x + y = `100sqrt3`

100 + y = 100 × 1.73

y = 173 − 100

y = 73 m

Time = `"Distance"/"Speed"`

= `(73  m)/(60  km//hr)`

= `(73 xx 60 xx 60)/(60 xx1000)`

= `(73 xx 6)/100`

= `438/100`

= 4.38 seс.

OR

b.

In ΔPQR

tan 60 = `(PQ)/(PR)`

`sqrt3 = 100/(PR)`

PR = `100/sqrt3`

Time = `"Distance"/"Speed"`

= `(100/(sqrt3 m))/(100/(6 m//sec))`

= `6/sqrt3`sec

= `6/sqrt3 xx sqrt3/sqrt3`

= `(6sqrt3)/3`

= `2sqrt3` sec.

= 2 × 1.73

= 3.46 sec.