Question

★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.
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Answer 1

Given :

• A man observes a car at an angle of depression of 30°.
• Six seconds later, the angle of depression of the car is found to be 60°.

To Find :

• Time taken by the car to reach the foot of the tower .

Solution :

$\longmapsto\tt{Let\: speed\: of\:car\:be=x}$

$\longmapsto\tt{Time\:taken\:by\:car\:CD=6\:sec}$

Now ,

$\longmapsto\tt{{Distance}_{CD}=Speed\times{time}}$

$\longmapsto\tt{x\times{6}}$

$\longmapsto\tt{6x\:sec}$

Similarly ,

$\longmapsto\tt{Time\:taken\:by\:car\:BC=t}$

Now ,

$\longmapsto\tt{{Distance}_{BC}=Speed\times{time}}$

$\longmapsto\tt{x\times{t}}$

$\longmapsto\tt{xt\:sec}$

$\longmapsto\tt{BD=BC+CD}$

$\longmapsto\tt\bf{xt+6x}$

In ∆ ABC :

$\longmapsto\tt{tan\:60\degree=\dfrac{h}{B}}$

$\longmapsto\tt{\sqrt{3}=\dfrac{h}{xt}}$

$\longmapsto\tt{h=xt\times{\sqrt{3}}}$

$\longmapsto\tt\bf{h=\sqrt{3}\:xt}$---(1)

In ∆ ABD :

$\longmapsto\tt{tan\:30\degree=\dfrac{h}{B}}$

$\longmapsto\tt{\dfrac{1}{\sqrt{3}}=\dfrac{h}{xt+6x}}$

$\longmapsto\tt{\sqrt{3}h=xt+6x}$

Putting Value of h from equation 1 :

$\longmapsto\tt{\sqrt{3}\:xt\times{\sqrt{3}}=xt+6x}$

$\longmapsto\tt{3\:xt=xt+6x}$

$\longmapsto\tt{3xt-xt=6x}$

$\longmapsto\tt{2{\not{x}}t=6{\not{x}}}$

$\longmapsto\tt{2t=6}$

$\longmapsto\tt{t=\cancel\dfrac{6}{2}}$

$\longmapsto\tt\bf{t=3\:sec}$

Answer 2

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.

Given :

• A man standing at the top of the tower observes a car at an angle of depression of 30°.
• Six seconds later, the angle of depression of the car is found to be 60°.

To find :

• Time taken by car to reach the foot of the tower?

Required Answer :

• Time taken by car to reach the foot of the tower is 3 seconds

Step by step explanation :

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

• Let speed of car be s and time taken to cover distance B to O be m
• Time taken by car to cover distance from O to C = 6 seconds

Using formula :

$\qquad\red\bigstar\:{\underline{\boxed{\bf{Distance = Speed\:\times\:Time}}}}$

➡ㅤㅤㅤDistance(OC) = s × 6

➡ㅤㅤㅤDistance(OC) = 6s

Also,

➡ㅤㅤㅤDistance(BO) = s × m

➡ㅤㅤㅤDistance(BO) = sm

Using formula :

$\qquad\red\bigstar\:{\underline{\boxed{\bf{tan\theta = \dfrac{Perpendicular}{Base}}}}}$

• In ∆ ABO :

➡ㅤㅤㅤtan60° = AB/BO

➡ㅤㅤㅤ√3 = x/sm

➡ㅤㅤㅤx = √3 sm

• In ∆ ABC :

➡ㅤㅤㅤtan30° = AB/BC

➡ㅤㅤㅤ1/√3 = AB/(BO + OC)

➡ㅤㅤㅤ1/√3 = x/(sm + 6s)

➡ㅤㅤㅤ√3 x = sm + 6sㅤㅤㅤㅤㅤㅤ[1]

Putting value of x in [1] :

➡ㅤㅤㅤ√3 × √3 sm = sm + 6s

➡ㅤㅤㅤ3 sm = sm + 6s

➡ㅤㅤㅤ3 sm - sm = 6s

➡ㅤㅤㅤ2 sm = 6s

By cancelling s from both sides :

➡ㅤㅤㅤ2m = 6

➡ㅤㅤㅤm = 6/2

➡ㅤㅤㅤm = 3 seconds

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

∴ The time taken by the car to reach the ㅤfoot of the tower is 3 seconds