Question

Tejas is standing at the top of a building and observes a car at an angle of depression of 30° as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to 60° and at that moment, the car is 25 m away from the building.

Based on the information given above, answer the following questions:

(i) What is the height of the building?   [1]

(ii) What is the distance between the two positions of the car?   [1]

(iii) (a) What would be the total time taken by the car to reach the foot of the building from the starting point?   [2]

OR

(iii) (b) What is the distance of the observer from the car when it makes an angle of 60°?   [2]

Answer 1

Let the height of building AB = h m

The car is first at point D, observed at an angle of depression is 30°.

After 6 seconds, the car moves to point C, observed at an angle of depression is 60°.

The distance CB = 25 m.

Since AB is vertical and BC is horizontal, triangle ABC is right angled at B.

(i) Height of the building:

 
In ΔABC,

`tan 60^circ = P/B = (AB)/(BC)`

`tan 60^circ = h/25`

`sqrt(3) = h/25`

`h = 25sqrt(3)  m`

∴ Height of the building is `25sqrt(3)  m`.

(ii) Distance between two position of the car (means DC)

 
In ΔABD,

`tan 30^circ = P/B = (AB)/(BD)`

`1/sqrt(3) = (25sqrt(3))/(BD)`

`BD = 25sqrt(3) xx sqrt(3)`

BD = 25 × 3

BD = 75

∴ DC = BD – BC

DC = 75 – 25

DC = 50 m

(iii) (a) Total time taken by the car to reach the foot of the building from the starting point:

Car covers Dc distance in 6 seconds.

Distance (DC) = 50 m, Time (T₁) = 6 sec

Speed = `"Distance"/"Time"`

Speed = `50/6`

Speed = `25/3`

Speed = 8.33 m/s

Now, distance (BC) = 25 m,

Speed = 8.33 m/s   ...[Given uniform speed]

Time (T₂) = (?)

Speed = `"Distance"/"Time"`

`8.33 = 25/"Time"`

Time = `25/8.33` = 3.00 sec

Time (T₂) =  3 sec

∴ Total time = T₁ (from D to C) + T₂ (from C to B)

= 6 sec + 3 sec

= 9 sec

OR

(iii) (b) Distance of the observer from the car when it makes an angle of 60°:

 
Distance AC is the line of sight from observer A to car at C.

In ΔABC,

`sin 60^circ = P/H = (AB)/(AC)`

`sin 60^circ = (25sqrt(3))/(AC)`

 `sqrt(3)/2 = (25sqrt(3))/(AC)`

`AC = (25sqrt(3) xx 2)/sqrt(3)`

AC = 25 × 2

AC = 50 m

∴ The required distance is 50 m.