Question

5. A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the
tower with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60°. Find the time taken by the car to reach the foot of the tower from this point.​

Answer 1

Answer:

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Answer 2

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In right ∆ ACD,

$tan \: {60}^{o} = \frac{ AD}{CD} \\ \sqrt{3} = \frac{AD}{CD} \\ AD = \sqrt{3} \: CD \: \: \: \: \: \: \: - - - - - (1)$

In right ∆ ABD,

$tan \: {30}^{o} = \frac{AD}{BD} \\ \frac{1}{ \sqrt{3} } = \frac{ AD}{BD} \\ \\ \frac{BD}{ \sqrt{3} } = AD \: \: \: \: - - - - - - (2)$

By comparing ( 1 ) and ( 2 )

$\sqrt{3} \: CD \: = \frac{BD}{ \sqrt{3} } \\3 \: CD = BD$

But ( BD = BC + CD )

$3 \: CD = CD + BC \\ BC = 3 \: CD - CD \\ BC = 2CD$

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→ Time taken to cover BC = 6 seconds

→ Time taken to cover $\frac{BC}{2}=\frac{6}{2}=3\:seconds$

→ So, Time taken to cover CD = 3 seconds

Hence , it takes 3 seconds to reach to the foot of the tower.

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