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 $1.5m$, 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 $3m$, and inclined at an angle of $60°$ to the ground. What should be the length of the slide in each case?

Answer 1

Step 1: Finding the length of the slide for younger children.

Let ABC be the slide for the children below the age of $5$ years and PQR be the slide for the elder children.

In $∆\mathrm{ABC},\mathrm{B}$ is the right angle and $C=30°$.

We know that, $\mathrm{sin\theta }=\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}$

$$\begin{gathered} \Rightarrow \sin {30}^o=\frac{AB}{AC} \\ \Rightarrow \frac{1}{2}=\frac{1.5}{AC} \\ \Rightarrow AC=1.5\times 2 \\ \Rightarrow AC=3m \end{gathered}$$

Step 2- Finding the length of the slide for elder children

$\begin{array}{rcl}\mathrm{Sin\theta }& =& \frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}\\ \mathrm{Sin}{60}^o& =& \frac{PQ}{PR}\\ \frac{\sqrt{3}}{2}& =& \frac{3}{PR}\\ PR& =& \frac{6}{\sqrt{3}}\\ PR& =& \frac{6}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ PR& =& \frac{6}{3}\times \sqrt{3}\\ PR& =& 2\sqrt{3}\\ PR& =& 2\times 1.73\\ PR& =& 3.46m\end{array}$

Hence, the length of slide for the children below the age of $5$ is $3m$, and for the elder children is $3.46m$.