$A$ 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 $30^{\circ}$ 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 $60^{\circ}$ to the ground. What should be the length of the slide in each case?
(N/A) It can be observed that $AC$ and $PR$ are the slides for younger and elder children respectively.
In $\triangle ABC$:
$\sin 30^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{1.5}{AC}$
$AC = 3 \, m$
In $\triangle PQR$:
$\sin 60^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR}$
$\frac{\sqrt{3}}{2} = \frac{3}{PR}$
$PR = \frac{6}{\sqrt{3}} = 2\sqrt{3} \, m \approx 3.46 \, m$
In $\triangle ABC$:
$\sin 30^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{1.5}{AC}$
$AC = 3 \, m$
In $\triangle PQR$:
$\sin 60^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR}$
$\frac{\sqrt{3}}{2} = \frac{3}{PR}$
$PR = \frac{6}{\sqrt{3}} = 2\sqrt{3} \, m \approx 3.46 \, m$
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