Question

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

Question:

⠀⠀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.

Stated:

• 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°

To Find:

• the time taken by the car to reach the foot of the tower from this point.

Solution:

★ Let us assume that,

• AB = h
• DC = x
• BC = y

★ In triangle ABC,

$\dashrightarrow \tt \dfrac{h}{y} = \tan ( 60) = \sqrt{3}$

$\dashrightarrow \tt h = \sqrt{3} y ---(1)$

★ In triangle ABD

$\dashrightarrow \tt \dfrac{h}{x + y} = \tan(30)= \dfrac{1}{\sqrt{3} }$

$\dashrightarrow \tt \sqrt{3} h = x + Y$

★ Putting the value of h from equation 1,

$\dashrightarrow \tt \sqrt{3} (\sqrt{3}y ) = x + y$

$\dashrightarrow \tt 3y = x + y$

$\dashrightarrow \tt 2y = x ---(2)$

★ As per the second statement,

• Let the uniform speed be U m/s

As we know that :

$\star {\boxed{\tt{Speed = \dfrac{Distance}{Time} }}}$

$\dashrightarrow \tt u = \dfrac{x}{6}$

$\dashrightarrow \tt x = 6u ---(3)$

★ From equation 2 and 3 we get,

$\dashrightarrow \tt 2y = 6u$

$\dashrightarrow \tt y = 3u$

★ Time taken,

$\dashrightarrow \tt BC = \dfrac{Distance}{Speed}$

$\dashrightarrow \tt BC = \dfrac{y}{u}$

$\dashrightarrow \tt BC = \dfrac{3u}{u}$

$\dashrightarrow \tt BC = 3 sec$

Hence:

• The time taken to travel the distance BC is 3 seconds

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