Q) 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?
(ii) What is the distance between the two positions of the car?
(iii) (a) What would be the total time taken by the car to reach the foot of the building from the starting point?
(iii) (b) What is the distance of the observer from the car when it makes an angle of 60°?

(Q 36 – 30/2/2 – CBSE 2026 Question Paper)

Ans: 

Let’s make the equivalent diagram to calculate different values:

Here, let’s consider AB is the building and the car initially at point D, and then after 6 minutes, it is at point C.

Given that the angle of depression from point A at the Tower top to the boat at Point D is 30⁰,

∴ The elevation angle from D to point A will be 30⁰

Similarly, given that the angle of depression from point A at the Tower top to the boat at Point C is 60⁰,

∴ The elevation angle from C to point A will be 60⁰.

Given that the distance from C to point B at the foot of the tower is 25 m..

Let’s consider the distance covered by the car from point D to point C is D.

(i) Height of the building

In Δ ABC,

tan 60⁰ =

∴ √ 3 =

∴ H = 25 √3 = 43.3 m

Therefore, the height of the buiding is 43.3 m.

(ii) Distance between two positions of the car:

Step 1: From the diagram, car is moving from point D to point C

Hence, we need to find the distance CD. 

Let this distance be D.

Step 2: Now in Δ ABD,

tan 30⁰ =

∴

∴ BD = H√ 3 

∵ H = 25 √3 m (from part (i) above)

∴ BD = (25 √3) √ 3  = 75 m

Step 3: Now from the diagram, we have: BD = BC + CD

∴ BC + CD = 75

∴ 25 + CD = 75

∴ CD = 75 – 25 = 50 m

Therefore, the distance between the two positions of the car is 50 m.

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

Given that the car covers distance CD (or 50 m distance) in 6 seconds

∴ Speed of the car = ≈ 8.33 m/s

∵ Total distance from the starting point (point D) to foot of building (point B) is given by BD

And we calculated BD = 75 m

∴ Time = = 9 seconds

Therefore, car would take total 9 seconds to reach the foot of the building from the starting point.

(iii)(b) Distance of observer from car when angle is 60⁰

(Here, we need to calculate length of AC)

In Δ ABC, cos 60⁰ =

∴

∴ AC = 2 x 25 = 50 m

Therefore, the distance of observer from car is 50 m when angle is 60⁰

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