Position of observer = ‘D’
Initial position of car = ‘A’
then angle of depression
= ∠DA = 30°
∴ In ΔACD, ∠C = 30°
⇒ tan 30 = \(\frac{CD}{AC}\)= \(\frac{1}{\sqrt{3}}\)
⇒ CD = \(\frac{AC}{\sqrt{3}}\) ……………..(1)
and after 12 seconds of time, position of car = ’B’
then angle of depression
= ∠XDB = 60° = ∠DBC
∴ In ΔBCD, ∠B = 60°
⇒ tan 60° = \(\frac{CD}{BC}\) = √3
⇒ CD = BC√3 ……………… (2)
∴ (1) = (2)
⇒ CD = \(\frac{AC}{\sqrt{3}}\) = BC √3 ⇒ AC = 3BC
Now from the figure
AC = AB + BC
⇒ 3BC = AB + BC ( ∵ AC = 3BC)
⇒ AB = 2BC
So the time taken to cover the distance \(\overline{AB}\) or the distance 2 \(\overline{BC}\) =12 seconds
∴ Time taken to cover \(\overline{BC}\) distance
= 6 sec (∵ \(\frac{12}{2}\) )
Then the time taken to approach the tower = time taken to cover the distance
\(\overline{AC}\) = 3\(\overline{BC}\) = 3(6) = 18 seconds that means it takes 6 more seconds (18 – 12) to reach tower.