(a) Length of the height is 2.732 m
(b) Length of the base is 5.466 m
Explanation:
An image is attached for reference.
(a)
In ΔAOB,
[tex]sin 30^o = \frac{AO}{AB} \\\\0.5 = \frac{AO}{2} \\\\AO = 1 m[/tex]
In ΔBGD,
[tex]sin 60^o = \frac{BG}{BD} \\\\0.866 = \frac{BG}{2} \\\\BG = 1.732 m[/tex]
According to the figure, BG = OE = 1.732 m
Height of the tent, AE = AO + OE
= 1 m + 1.732 m
= 2.732 m
(b)
DF = ?
In ΔAOB,
[tex]tan 30^o = \frac{AO}{OB} \\\\0.577 = \frac{1}{OB} \\\\OB = 1.733 m\\\\\\[/tex]
According to the figure, OB = GE = 1.733 m
In ΔBGD,
[tex]tan 60^o = \frac{BG}{DG} \\\\1.732 = \frac{1.732}{DG}\\ \\DG = 1m[/tex]
According to the figure, DE = DG + GE
DE = 1 m + 1.733 m
DE = 2.733 m
Length of the base, DF = 2 X DE
DF = 2 X 2.733 m
DF = 5.466 m
