NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 6
NCERT Class X Chapter 9: Some Applications of Trigonometry
Question:
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Given:
- Height of the boy = 1.5 m
- Height of the building = 30 m
- Initial angle of elevation = 30°
- Final angle of elevation = 60°
To Find:
The distance the boy walked towards the building.
Formula:
\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \]
Solution:
Step 1: Let the initial distance from the boy to the building be \( x \) metres, and the final distance be \( y \) metres. The effective height from the boy's eyes to the top of the building is:
\[ \text{Effective height} = 30\,\text{m} - 1.5\,\text{m} = 28.5\,\text{m} \]Step 2: Using the tangent ratio for the initial position (angle of elevation \( 30^\circ \)):
\[ \tan 30^\circ = \frac{28.5}{x} \] \[ x = \frac{28.5}{\tan 30^\circ} \]Step 3: Using the tangent ratio for the final position (angle of elevation \( 60^\circ \)):
\[ \tan 60^\circ = \frac{28.5}{y} \] \[ y = \frac{28.5}{\tan 60^\circ} \]Step 4: Substitute the values of \( \tan 30^\circ = \frac{1}{\sqrt{3}} \) and \( \tan 60^\circ = \sqrt{3} \):
\[ x = \frac{28.5}{\frac{1}{\sqrt{3}}} = 28.5 \times \sqrt{3} \] \[ y = \frac{28.5}{\sqrt{3}} \]Step 5: Calculate the values:
\[ x = 28.5 \times 1.732 \approx 49.32\,\text{m} \] \[ y = \frac{28.5}{1.732} \approx 16.46\,\text{m} \]Step 6: The distance walked towards the building is:
\[ x - y = 49.32\,\text{m} - 16.46\,\text{m} = 32.86\,\text{m} \]Result:
The distance the boy walked towards the building is approximately \( 32.9 \) metres.
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