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A man on the top of the vertical tower observes a car moving a

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A man on the top of the vertical 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 its measure from $30$ to $45$ , how soon after this, will the car reach the tower ?

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A man on the top of the vertical 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 its measure from

30

45

, how soon after this, will the car reach the tower ?

5 mins ago

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Text solutionVerified

Let speed of can be

v m / min

\Rightarrow BC = v \times 12

Δ A OC, tan 30 = \frac{A O}{OC} \Rightarrow \frac{1}{3} = \frac{A O}{OC}

Δ A OB, tan 45 = \frac{A O}{OB} \Rightarrow 1 = \frac{A O}{OB}

\Rightarrow OC = 3 A O = 3 OB

\Rightarrow OB + BC = 3 OB

\Rightarrow OB (3 - 1) \Rightarrow BC \Rightarrow \frac{OB}{BC} = \frac{1}{3 - 1}

Now,

t = \frac{OB}{v} = \frac{BC}{v ( 3 - 1 )} = \frac{12}{3 - 1} \times \frac{3 + 1}{3 + 1}

t = 6 (3 + 1) min

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Question Text	A man on the top of the vertical 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 its measure from $30$ to $45$ , how soon after this, will the car reach the tower ?
Answer Type	Text solution:1
Upvotes	150