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 $30^o$ 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 $60^o$ to the ground. What should be the length of the slide in each case?

Given:

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 $30^o$ 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 $60^o$ to the ground. 

To do:

We have to find the length of the slide in each case.

Solution:  

Let $AB$ be the height of the slide above the ground and $AC$ be the length of the slide for children below the age of 5 years.

Let $PQ$ be the height of the slide above the ground and $PR$ be the length of the slide for children above the age of 5 years.

From the figures,

In $\triangle ABC$,

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

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

$\mathrm{AC}=3 \mathrm{~m}$

In $\triangle PQR$,

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

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

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

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

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

$=2 \sqrt{3} \mathrm{~m}$

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Updated on: 10-Oct-2022

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