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Answer :
Let AB be the tower of height h. Let C be the initial position of the car and let after 6 seconds, the car be at D. It is given that the angles of depression at C and D are 30º and 60º respectively. Let the speed of the car be v m/s. Then ,
CD = Distance travelled by the car in 6 seconds
\(=> CD \ = \ 6v \) metres
[\(\because \) Distance = Speed × Time]
Let the car takes t seconds to reach the tower AB from C. Then,
\( DA \ = \ vt \) metres
In \( ∆ \ ABC \), we have
\( \frac{AB}{AC} \ = \ tan60º \)
\( => \frac{h}{vt} \ = \ \sqrt{3} \)
\( => \ h \ = \sqrt{3} vt \) (1)
In \( ∆ \ ABD \) , we have
\( \frac{AB}{AD} \ = \ tan30º \)
\( => \frac{h}{vt + 6v} \ = \frac{1}{\sqrt{3}} \)
\( => \ h\sqrt{3} \ = vt + 6v \) .... (2)
Substituting the value of h from (1) in (2), we get
\( \sqrt{3} \ × \ \sqrt{3}vt \ = \ vt + 6t \)
\( => 3vt \ = \ vt \ + \ 6v \)
\( => \ 3t \ - \ t \ = \ 6 \)
\( => t \ = \ 3 \)
\(\therefore \) the car will reach the tower from C in 3 seconds.