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in Trigonometry by (41.6k points)
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A man on the top of vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 seconds to change the angle of depression from 30° to 60°, then how long will the car take to reach the tower from that point.

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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.

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