Question

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आप से ये निवेदन है की क्या आप इस गणितीय सवाल का उत्तर अंग्रेजी में दे सकते हैं।❓

A straight highway leads to the foot of a tower. Rahul standing at the top of the tower observes a car at an angle of depression 30°. The car is approaching the foot of the tower with a uniformspeed. 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

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⎟⎟✪✪ प्रश्न ✪✪⎟⎟

A straight highway leads to the foot of a tower. Rahul standing at the top of the tower observes a car at an angle of depression 30°. The car is approaching the foot of the tower with a uniformspeed. 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.

⎟⎟✪✪ उत्तर ✪✪⎟⎟

◆ Let the distance travelled by the car in 6 seconds = AB = x meters

◆ Height of the tower CD = h meters

◆ The remaining distance to be travelled by the car BC = d meters

And AC = AB + BC = (x + d) meters

∠PDA = ∠DAC = 30°

∠PDB = ∠DBC = 60°

From ΔBCD

$\implies$tan60° = CD/BC

$\implies$√3 = h/d

$\implies$h = √3d ━━━▶➀

From ΔACD

$\implies$tan30° = CD/AC

$\implies$1/√3 = h/(x + d)

$\implies$h = (x + d)/√3 ━━━▶➁

From ➀ and ➁, we have

$\implies$x+d/√3 = √3d

$\implies$x+d = 3d

$\implies$x = 2d

$\implies$ d = x/2

$\huge\rightarrow$Time taken to travel 'x'meters = 6 seconds

$\huge\rightarrow$Time taken to travel the distance of 'd'meters

i.e., x/2 meters = 3 seconds

Answer 2

Answer:

in triangle ABC

tan 60 = h/t

√3/1 = h/t

√3/1 × h/t

h= √3t. eq-1

in triangle ABD

tan 30 = h/6sec

1/√3 × h/6sec

√3h = 6sec. eq-2

from eq1 and 2

√3×√3t = 6sec

3t - t = 6sec

2t = 6sec

t = 3sec answer