Mathematics Question and Solution
tan15 = a/4
a = 1.07179676972 cm.
Where a is AD.
b² = 1.07179676972²+4²
b = 4.14110472164 cm.
Where b is DF.
c = 4-1.07179676972
c = 2.92820323028 cm.
Where c is CD.
sin60 = d/2.92820323028
d = 2.53589838487 cm.
Where d is DE = BE.
e² = 16+16-2*16cos150
e = 7.72740661031 cm.
Where e is BF.
f² = 7.72740661031²+2.53589838487²-2*2.53589838487*7.72740661031cos45
f = 6.19925645689 cm.
Where f is EF.
(6.19925645689/sin45) = (2.53589838487/sing)
g = 16.81321456806°
Where g is angle BFE.
h = 15+g
h = 31.81321456806°
Where h is angle AFE.
Therefore;
Length FG is;
cos31.81321456806 = 4/(FG)
FG = 4/(cos31.81321456806)
FG = 4.70714976963 cm.
