Question

A carpenter makes stools for electricians with a square top of side 0.5m and at a height 1.5m above the ground. Also, each leg is inclined at an angle of ${60}^{\circ }$ to the ground. Find the lengths of two steps in metres correct up to two places of decimals.

Answer 1

As shown in the figure, AX and BY are the two legs at an angle of ${60}^{\circ }$ to the ground. CD and EF are two steps. AR and BS are perpendiculars on the base XY.

AR meets CD at L and EF at P.
BS meets CD at M and EF at Q.
We have, AR = BS = 1.5 m, AB = 0.5 m
AL = LP = PR = BM = MQ = QS = 0.5 m
From right angled $\Delta AXR$,
$sin{60}^{\circ }=\frac{AR}{AX}\Rightarrow \frac{\sqrt{3}}{2}=\frac{1.5}{AX}\Rightarrow AX=\frac{1.5\times 2}{\sqrt{3}}=\frac{3}{\sqrt{3}}\Rightarrow AX=\sqrt{3}m\Rightarrow AX=1.73m\Rightarrow AX=BY=1.73m$

In right angled $\Delta ACLand\Delta AEP,\angle ACL={60}^{\circ }and\angle AEP={60}^{\circ }$
Then $\frac{AL}{CL}=tan{60}^{\circ }and\frac{AP}{EP}=tan{60}^{\circ }\Rightarrow \frac{0.5}{CL}=\sqrt{3}and\frac{1}{EP}=\sqrt{3}\Rightarrow CL=\frac{0.5}{\sqrt{3}}andEP=\frac{1}{\sqrt{3}}\Rightarrow CL=\frac{0.5\times \sqrt{3}}{3}andEP=\frac{\sqrt{3}}{3}$
Similarly, $MD=\frac{0.5\times \sqrt{3}}{3}andQF=\frac{\sqrt{3}}{3}$
Now, CD = CL + LM + MD
$=\frac{0.5\times \sqrt{3}}{3}+0.5+\frac{0.5\times \sqrt{3}}{3}=0.5+\frac{\sqrt{3}}{3}=0.5+\frac{1.73}{3}$
= 0.5 + 0.577 (approx) = 1.077 m

Again, $EF=EP+PQ+QF=\frac{\sqrt{3}}{3}+0.5+\frac{\sqrt{3}}{3}=0.5+\frac{2\sqrt{3}}{3}=0.5+\frac{2\times 1.73}{3}=0.5+\frac{3.46}{3}=0.5+1.153=1.623m$

Hence, and length of step CD = 1.077 m and length of step EF = 1.653 m