NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 12

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Given:

  • Height of the building = 7 m
  • Angle of elevation to top of tower = 60°
  • Angle of depression to foot of tower = 45°

To Find:

Height of the tower

Formula:

\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \]

Solution:

Step 1: Let the height of the building be \( AB = 7\,\text{m} \). Let the height of the tower be \( CD = h\,\text{m} \). Let the horizontal distance between the building and the tower be \( x\,\text{m} \).

Step 2: From the top of the building, the angle of depression to the foot of the tower is 45°. In right triangle \( ABE \):

$$ \tan 45^\circ = \frac{AB}{BE} $$ $$ 1 = \frac{7}{x} \implies x = 7\,\text{m} $$

Step 3: The angle of elevation to the top of the tower from the building is 60°. In right triangle \( DAE \):

$$ \tan 60^\circ = \frac{h - 7}{x} $$ $$ \sqrt{3} = \frac{h - 7}{7} $$ $$ h - 7 = 7\sqrt{3} $$

Step 4: Solving for \( h \):

$$ h = 7 + 7\sqrt{3} $$ $$ h \approx 7 + 7 \times 1.732 = 7 + 12.124 = 19.124\,\text{m} $$

Result:

The height of the tower is approximately \( 19.12\,\text{m} \).

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