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Question
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.
Solution
In ΔPSB,
`(PS)/(PB) = sin 60^circ`
`=> PB = 2/sqrt(3) = 1.155 m`
In ΔAPQ,
∠APQ = 60°
∴ `(PQ)/(AP) = cos 60^circ`
`=> AP = 1/(1/2) = 2 m`
∴ AB = AP + PB
= 2 + 1.155
= 3.155 m
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