NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 3

NCERT Class X Chapter 9: Some Applications of Trigonometry

Question:

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Given:

  • Height of slide for children below 5 years, \( h_1 = 1.5\,\text{m} \)
  • Angle of inclination for children below 5 years, \( \theta_1 = 30^\circ \)
  • Height of slide for elder children, \( h_2 = 3\,\text{m} \)
  • Angle of inclination for elder children, \( \theta_2 = 60^\circ \)

To Find:

  • Length of the slide for children below 5 years (\( l_1 \))
  • Length of the slide for elder children (\( l_2 \))

Formula:

\[ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} \]

Solution:

Step 1: Let the length of the slide for children below 5 years be \( l_1 \). Using the formula,

\[ \sin 30^\circ = \frac{h_1}{l_1} \] \[ \sin 30^\circ = \frac{1.5}{l_1} \]

Step 2: Substitute the value of \( \sin 30^\circ = \frac{1}{2} \) and solve for \( l_1 \).

\[ \frac{1}{2} = \frac{1.5}{l_1} \] \[ l_1 = \frac{1.5}{\frac{1}{2}} = 1.5 \times 2 = 3\,\text{m} \]

Step 3: Let the length of the slide for elder children be \( l_2 \). Using the formula,

\[ \sin 60^\circ = \frac{h_2}{l_2} \] \[ \sin 60^\circ = \frac{3}{l_2} \]

Step 4: Substitute the value of \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and solve for \( l_2 \).

\[ \frac{\sqrt{3}}{2} = \frac{3}{l_2} \] \[ l_2 = \frac{3}{\frac{\sqrt{3}}{2}} = 3 \times \frac{2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}\,\text{m} \] \[ 2\sqrt{3} \approx 2 \times 1.732 = 3.464\,\text{m} \]

Result:

Length of slide for children below 5 years, \( l_1 = 3\,\text{m} \)
Length of slide for elder children, \( l_2 \approx 3.46\,\text{m} \)

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