Q) A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of 30 deg who is approaching the foot of the tower with a uniform speed. 30 seconds later, the angle of depression changes to 60 deg. Find the time taken by his friend to reach the foot of the tower from this point.

(Q 34B – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s draw the diagram for better understanding of the question:

Here, A is the man on tower AB. Let’s consider height as H. His friend is at point C with angle of depression (AOD) as 30 deg. After 30 secs, the friend moves P distance, reaches at Point D and now AOD is 60 deg. We need to find time to travel distance Q (from D to B)

Next, We plot corresponding alternate angles to AOD.

Step 2: In Δ ABC, tan 30 =

∴

∴ P + Q = H √3 …………(i)

Step 2: Now, In Δ ABD, tan 60 =

∴ √3 =  

∴ H = Q √3 …………. (ii)

Step 3: By solving equation (i) & equation (ii), we get:

∴ P + Q = (Q√3) √3

∴ P + Q = 3 Q

∴ P = 3 Q – Q= 2 Q

∴ P = 2 Q ……….. (iii)

Step 4: Let’s consider the moving friend’s speed is V and time taken to move P distance is T₁

∵ Distance = speed x time

∴ P = V x T₁

It is given that time taken was 30 secs

∴ P = V x 30 = 30 V

∵ From equation (iii), we have P = 2 Q

∴ 2 Q = 30 V

∴ Q = 15 V ……. (iv)

Step 5: Now friend needs to travel distance Q at same speed V.

Let’s consider time taken to move Q distance is T₂

∴ Q = V x T₂

∵ From equation (iv), we have Q= 15 V

∴ 15 V = V x T₂

∴ T₂ =

∴ T₂ = 15

Therefore, time taken to reach the foot of the tower from D will be 15 secs.

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