In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.

In ∆PSB,

$\text{sin 60°} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{PS}{PB} \\[1em] \Rightarrow PB = \dfrac{2PS}{\sqrt{3}} \\[1em] \Rightarrow PB = \dfrac{2 \times 1}{1.732} = 1.155 \text{ m}.$

In ∆APQ,

∠APQ = ∠ABR = 60° (Corresponding angles are equal.)

$\text{cos 60°} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{PQ}{AP} \\[1em] \Rightarrow AP = 2PQ \\[1em] \Rightarrow AP = 2 \text{ m}.$

From figure,

AB = AP + PB = 2 + 1.155 = 3.155 m.

Hence, AB = 3.155 meters.