A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point?

Question 27 (OR 1st question)
A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how long will the car take to reach the observation tower from this point?
Let the man & tower height be AD
Given that man sees car first at an angle of depression of 30°
So, ∠ PAB = 30°
After 12 minutes, the man sees car at an angle of depression of 45°
So, ∠ PAC = 45°
Now, tower is vertical,
So, ∠ ADB = 90°
Also,
Lines PA & BD are parallel
And AB is the transversal
∠ ABD = ∠ PAB (Alternate Angles)
So, ∠ ABD = 30°
Lines PA & BD are parallel
And AC is the transversal
∠ ACD = ∠ PAC (Alternate Angles)
So, ∠ ACD = 45°
In right angle triangle ACD
tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶)
tan 45° = AD/CD
1 = 𝐴𝐷/𝐶𝐷
CD = AD
AD = CD
In right angle triangle ABD
tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)
tan 30° = (" " AD)/BD
(" " 1)/√3 = 𝐴𝐷/𝐵𝐷
BD/√3 = AD
AD = BD/√3
From (1) & (2)
CD = 𝐵𝐷/√3
√3CD = BD
√3CD = BC + CD
√3CD – CD = BC
CD (√3 – 1) = BC
CD = 𝐵𝐶/((√3 − 1) )
CD = 𝐵𝐶/((√3 − 1) ) × ((√3 + 1))/((√3 + 1) )
CD = 𝐵𝐶(√3 + 1)/(((√3)^2 − 1^2 ) )
CD = 𝐵𝐶(√3 + 1)/((3 − 1) )
CD = 𝐵𝐶(√3 + 1)/2
Given that
Time taken to cover BC = 12 minutes
Time taken to cover 𝐵𝐶(√3 + 1)/2 = 12 × ((√3 + 1))/2 minutes
= 6 (√3 + 1) minutes
∴ Time taken to cover CD = 6 (√3 + 1) minutes
Hence, it takes 6 (√𝟑 + 1) minutes to reach to the foot of the tower.

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.