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?

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Solution

## 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}}$$⇒\mathrm{sin}{30}^{o}=\frac{AB}{AC}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{1.5}{AC}\phantom{\rule{0ex}{0ex}}⇒AC=1.5×2\phantom{\rule{0ex}{0ex}}⇒AC=3m$Step 2- Finding the length of the slide for elder children$\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}}×\frac{\sqrt{3}}{\sqrt{3}}\\ PR& =& \frac{6}{3}×\sqrt{3}\\ PR& =& 2\sqrt{3}\\ PR& =& 2×1.73\\ 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 $3.46m$.

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