A carpenter makes stools for electricians with a square top of side $ 0.5 \mathrm{~m} $ and at a height of $ 1.5 \mathrm{~m} $ above the ground. Also, each leg is inclined at an angle of $ 60^{\circ} $ to the ground. Find the length of each leg and also the lengths of two steps to be put at equal distances.

Given:

A carpenter makes stools for electricians with a square top of side \( 0.5 \mathrm{~m} \) and at a height of \( 1.5 \mathrm{~m} \) above the ground. Also, each leg is inclined at an angle of \( 60^{\circ} \) to the ground.

To do:

We have to find the length of each leg and also the lengths of two steps to be put at equal distances.

Solution:

Let $AC$ be the leg of the stool whose top is a square-shaped of side $AB$.

Height of the stool $AS = 1.5\ m$ and the angle of inclination of the leg of the stool is $60^{\circ}$.
Let $AC=x\ m$

In right $\Delta \mathrm{ACS}$,

$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

$=\frac{\mathrm{AS}}{\mathrm{AC}}$

$\Rightarrow \sin 60^{\circ}=\frac{1.5}{x}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{2 x}$ $\Rightarrow 2\sqrt{3} x=3 \times 2$

$\Rightarrow x=\frac{6}{2 \sqrt{3}}$

$=\sqrt{3}$

$=1.732$

The length of the leg is $1.732 \mathrm{~m}$.

Given that there are two steps on equal distance.

This implies,

The distance between two steps $=\frac{1.5}{3}=0.5 \mathrm{~m}$

From the figure,

$\mathrm{EF}\|\mathrm{GH}\| \mathrm{CD}$ $\angle \mathrm{E}=\angle \mathrm{G}=\angle \mathrm{C}=60^{\circ}$        (Corresponding angles are equal)

In $\Delta \mathrm{AGT}$,

$\tan 60^{\circ}=\frac{\mathrm{AT}}{\mathrm{GT}}$

$\Rightarrow \sqrt{3}=\frac{1}{\mathrm{GT}}$

$\Rightarrow \mathrm{GT}=\frac{1}{\sqrt{3}}$

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

$=\frac{1.732}{3}$

$=0.577\ m$

$\mathrm{GH}=0.5+0.577+0.577=1.645 \mathrm{~m}$

Similarly,

In $\triangle \mathrm{AEU}$,

$\tan 60^{\circ}=\frac{\mathrm{AU}}{\mathrm{EU}}$

$\Rightarrow \sqrt{3}=\frac{0.5}{\mathrm{EU}}$

$\Rightarrow \mathrm{EU}=\frac{0.5}{\sqrt{3}}$

$=\frac{1}{2 \sqrt{3}}$

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

$=\frac{1.732}{6}$

$=0.288$

$\mathrm{EF}=0.5+0.2886+0.2886=1.077 \mathrm{~m}$

The length of each leg is $1.732\ m$ and the lengths of two steps to be put at equal distances are $1.1077\ m$ and $1.654\ m$.

Updated on: 10-Oct-2022

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