Question

2. A square-based tent has the cross-sectional

Answer 1

(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,

$sin 30^o = \frac{AO}{AB} \\\\0.5 = \frac{AO}{2} \\\\AO = 1 m$

In ΔBGD,

$sin 60^o = \frac{BG}{BD} \\\\0.866 = \frac{BG}{2} \\\\BG = 1.732 m$

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,

$tan 30^o = \frac{AO}{OB} \\\\0.577 = \frac{1}{OB} \\\\OB = 1.733 m$

According to the figure, OB = GE = 1.733 m

In ΔBGD,

$tan 60^o = \frac{BG}{DG} \\\\1.732 = \frac{1.732}{DG}\\ \\DG = 1m$

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