Math, asked by pravdeep3187, 11 months ago

A straight highway leads to the foot of a tower. Ramaiah standing at the top of the tower observes a car at an angle of depression 30º. The car 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 60º. Find the time taken by the car to reach the foot of the tower from this point.

Answers

Answered by Anonymous
1

Answer:

3 seconds

Step-by-step explanation:

Angle of depression = 30º (Given)

Angle of depression after 6 seconds = 60º (Given)

Let the distance traveled by the car in 6 seconds = AB = x m

Let the height of the tower CD = h m

Thus, the remaining distance to be traveled by the car BC = d m

AC = AB+BC = x + d

∠PDA = ∠DAC = 30°

∠ PDB = ∠DBC = 60°

In Δ BCD

TanФ = CD/BC

Tan60 =CD/BC

√3 =h/d

h = √3d

In Δ ACD

TanФ = CD/AC

Tan30 =CD/AC

1/√3 = h/(x+d)

h =(x+d)/√3

(x+d)/√3 =√3d

x+d = 3d

x =3d - d

x = 2d

d =x/2

Thus, time taken to travel x m = 6 secs

Time taken to travel the distance of D metres = x/2 = 6/2 = 3

Therefore, the time taken by the car to reach the foot of the tower from this point is 3 seconds.

Answered by Anonymous
1

Answer:

Step-by-step explanation:

Angle of depression = 30º (Given)

Angle of depression after 6 seconds = 60º (Given)

Let the distance traveled by the car in 6 seconds = AB = x m

Let the height of the tower CD = h m

Thus, the remaining distance to be traveled by the car BC = d m

AC = AB+BC = x + d

∠PDA = ∠DAC = 30°

∠ PDB = ∠DBC = 60°

In Δ BCD

TanФ = CD/BC

Tan60 =CD/BC

√3 =h/d

h = √3d

In Δ ACD

TanФ = CD/AC

Tan30 =CD/AC

1/√3 = h/(x+d)

h =(x+d)/√3

(x+d)/√3 =√3d

x+d = 3d

x =3d - d

x = 2d

d =x/2

Thus, time taken to travel x m = 6 secs

Time taken to travel the distance of D metres = x/2 = 6/2 = 3

Therefore, the time taken by the car to reach the foot of the tower from this point is 3 seconds.

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