NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 13

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Given:

  • Height of lighthouse = 75 m
  • Angle of depression of first ship = 45°
  • Angle of depression of second ship = 30°
  • Both ships are on the same side and one is exactly behind the other

To Find:

Distance between the two ships.

Formula:

The tangent of an angle in a right triangle is given by:
$$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$

Solution:

Step 1: Let the distance of the first ship from the base of the lighthouse be \( x_1 \) and the distance of the second ship be \( x_2 \), where the second ship is farther from the lighthouse than the first.

Step 2: For the first ship (angle of depression = 45°), apply the tangent formula:

$$ \tan 45^\circ = \frac{75}{x_1} $$

Since \( \tan 45^\circ = 1 \):

$$ 1 = \frac{75}{x_1} \implies x_1 = 75 \text{ m} $$

Step 3: For the second ship (angle of depression = 30°), apply the tangent formula:

$$ \tan 30^\circ = \frac{75}{x_2} $$

Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):

$$ \frac{1}{\sqrt{3}} = \frac{75}{x_2} $$ $$ x_2 = 75 \sqrt{3} \text{ m} $$

Step 4: Find the distance between the two ships:

$$ \text{Distance} = x_2 - x_1 = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) \text{ m} $$

Result:

The distance between the two ships is $$75(\sqrt{3} - 1)$$ m.

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