NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 4

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Given:

Angle of elevation, \( \theta = 30^\circ \)
Distance from the point to the foot of the tower, \( = 30\,\text{m} \)

To Find:

Height of the tower (\( h \))

Formula:

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

Solution:

Step 1: Let the height of the tower be \( h \) m. The distance from the point to the foot of the tower is 30 m.

Step 2: Using the definition of tangent for the angle of elevation,

$$ \tan 30^\circ = \frac{h}{30} $$

Step 3: We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). Substitute this value:

$$ \frac{1}{\sqrt{3}} = \frac{h}{30} $$

Step 4: Cross-multiply to solve for \( h \):

$$ h = 30 \times \frac{1}{\sqrt{3}} = \frac{30}{\sqrt{3}} $$

Step 5: Rationalise the denominator:

$$ h = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} $$

Result:

The height of the tower is \( 10\sqrt{3} \) m.

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