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Q3:
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?

Solution :

Given:

Case 1 (Children below 5 years):

  • Height of the slide ($AB$) = $1.5$ m
  • Angle of inclination ($\angle ACB$) = $30^\circ$

Case 2 (Elder children):

  • Height of the slide ($PQ$) = $3$ m
  • Angle of inclination ($\angle PRQ$) = $60^\circ$

To find:

The length of the slide for each case (i.e., the length of the hypotenuse $AC$ and $PR$).

1.5m AC 30° 3m PR 60°

Step 1: Calculating the length of the slide for children below 5 years.

In the right-angled triangle $\triangle ABC$, where $\angle B = 90^\circ$:

We use the trigonometric ratio that relates the opposite side to the hypotenuse, which is the sine function:

$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$

Substituting the given values:

$\sin(30^\circ) = \frac{AB}{AC}$

Since $\sin(30^\circ) = \frac{1}{2}$ [Standard trigonometric value]:

$\frac{1}{2} = \frac{1.5}{AC}$

By cross-multiplication:

$AC = 1.5 \times 2$

$AC = 3$ m

Step 2: Calculating the length of the slide for elder children.

In the right-angled triangle $\triangle PQR$, where $\angle Q = 90^\circ$:

Using the sine function again:

$\sin(60^\circ) = \frac{PQ}{PR}$

Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ [Standard trigonometric value]:

$\frac{\sqrt{3}}{2} = \frac{3}{PR}$

By cross-multiplication:

$PR \times \sqrt{3} = 3 \times 2$

$PR \times \sqrt{3} = 6$

$PR = \frac{6}{\sqrt{3}}$

Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

$PR = \frac{6 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$PR = \frac{6\sqrt{3}}{3}$

$PR = 2\sqrt{3}$ m

Using the approximation $\sqrt{3} \approx 1.732$:

$PR \approx 2 \times 1.732 = 3.464$ m

Final Answer: The length of the slide for children below 5 years is $3$ m, and the length of the slide for elder children is $2\sqrt{3}$ m (approximately $3.46$ m).


More Questions from Class 10 Mathematics Applications of Trigonometry EXERCISE 9.1


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