NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 15

NCERT Class X Chapter 9: Some Applications of Trigonometry

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.

Given:

  • Angle of depression at first observation = \(30^\circ\)
  • Angle of depression at second observation = \(60^\circ\)
  • Time between observations = 6 seconds
  • The car moves with uniform speed

To Find:

Time taken by the car to reach the foot of the tower from the second observation point.

Formula:

  • \(\tan \theta = \dfrac{\text{Opposite side}}{\text{Adjacent side}}\)
  • Uniform speed: \(\dfrac{\text{Distance}_1}{\text{Time}_1} = \dfrac{\text{Distance}_2}{\text{Time}_2}\)

Solution:

Step 1: Let the height of the tower be \( h \). Let the distance of the car from the foot of the tower at the first observation be \( x_1 \), and at the second observation be \( x_2 \).

Step 2: At the first observation, the angle of depression is \(30^\circ\). Using trigonometry:

$$ \tan 30^\circ = \frac{h}{x_1} \implies x_1 = h \sqrt{3} $$

Step 3: At the second observation, the angle of depression is \(60^\circ\):

$$ \tan 60^\circ = \frac{h}{x_2} \implies x_2 = \frac{h}{\sqrt{3}} $$

Step 4: The distance covered by the car in 6 seconds is:

$$ x_1 - x_2 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = h\left( \frac{3 - 1}{\sqrt{3}} \right) = h\left( \frac{2}{\sqrt{3}} \right) $$

Step 5: The remaining distance to the tower from the second observation point is:

$$ x_2 = \frac{h}{\sqrt{3}} $$

Step 6: Since the speed is uniform, the ratio of distances equals the ratio of times. Let \( t \) be the required time:

$$ \frac{x_1 - x_2}{x_2} = \frac{6}{t} $$ $$ \frac{h\left( \frac{2}{\sqrt{3}} \right)}{\frac{h}{\sqrt{3}}} = \frac{6}{t} $$ $$ \frac{2}{1} = \frac{6}{t} $$

Step 7: Solve for \( t \):

$$ 2 = \frac{6}{t} \implies t = \frac{6}{2} = 3 $$

Result:

The time taken by the car to reach the foot of the tower from the second observation point is 3 seconds.

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