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Hint: Consider a different right angled triangle for both the cases. Use a suitable trigonometric ratio to evaluate the value.

For slides of younger children, consider the first triangle, $\Delta ABC$.

AC is the height of the top of the slide from the ground and it is 1.5 m (given in the question).

Angle of inclination of the slide to the ground is ${30^ \circ }$ so $\angle ABC = {30^ \circ }$. We have to calculate the length of the slide i.e. BC. So in $\Delta ABC$:

$

\Rightarrow \sin {30^ \circ } = \dfrac{{AC}}{{BC}}, \\

\Rightarrow \dfrac{1}{2} = \dfrac{{1.5}}{{BC}}, \\

\Rightarrow BC = 3 \\

$

Thus, the length of the slide in this case is 3 m.

Next for slides of younger children, consider $\Delta XYZ$

YZ is the height of the top of the slide from the ground and it is 3 m (given in the question).

Angle of inclination of the slide to the ground is ${60^ \circ }$ so $\angle YXZ = {60^ \circ }$. We have to calculate the length of the slide i.e. XZ. So in $\Delta XYZ$:

$

\Rightarrow \sin {60^ \circ } = \dfrac{{YZ}}{{XZ}}, \\

\Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{3}{{XZ}}, \\

\Rightarrow XZ = \dfrac{{3 \times 2}}{{\sqrt 3 }}, \\

\Rightarrow XZ = 2\sqrt 3 \\

$

Thus, the length of the slide in this case is $2\sqrt 3 $ m.

Note: In the above two scenarios, perpendicular was known to us and we were required to calculate hypotenuse. Trigonometric ratio concerning these two sides is $\sin \theta $. That’s why we used it. In different scenarios, we can use different trigonometric ratio as per the convenience.

For slides of younger children, consider the first triangle, $\Delta ABC$.

AC is the height of the top of the slide from the ground and it is 1.5 m (given in the question).

Angle of inclination of the slide to the ground is ${30^ \circ }$ so $\angle ABC = {30^ \circ }$. We have to calculate the length of the slide i.e. BC. So in $\Delta ABC$:

$

\Rightarrow \sin {30^ \circ } = \dfrac{{AC}}{{BC}}, \\

\Rightarrow \dfrac{1}{2} = \dfrac{{1.5}}{{BC}}, \\

\Rightarrow BC = 3 \\

$

Thus, the length of the slide in this case is 3 m.

Next for slides of younger children, consider $\Delta XYZ$

YZ is the height of the top of the slide from the ground and it is 3 m (given in the question).

Angle of inclination of the slide to the ground is ${60^ \circ }$ so $\angle YXZ = {60^ \circ }$. We have to calculate the length of the slide i.e. XZ. So in $\Delta XYZ$:

$

\Rightarrow \sin {60^ \circ } = \dfrac{{YZ}}{{XZ}}, \\

\Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{3}{{XZ}}, \\

\Rightarrow XZ = \dfrac{{3 \times 2}}{{\sqrt 3 }}, \\

\Rightarrow XZ = 2\sqrt 3 \\

$

Thus, the length of the slide in this case is $2\sqrt 3 $ m.

Note: In the above two scenarios, perpendicular was known to us and we were required to calculate hypotenuse. Trigonometric ratio concerning these two sides is $\sin \theta $. That’s why we used it. In different scenarios, we can use different trigonometric ratio as per the convenience.

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