Mathematics Question and Solution
a = acos(3/5)
Where a is angle GCA.
tan(acos(3/5)) = b/3
b = 4 units.
Where b is AG.
c = atan(⅔)
Where c is angle GBE.
d = 180-acos(3/5)-atan(⅔)
d = 93.17983011986°
Where d is angle BEC.
(6/sin93.17983011986 ) = (e/sin(acos(⅗)))
e = 4.80740170062 units.
Where e is BE.
Area Parallelogram BCDE is;
2(area triangle with height 4.80740170062 units and base 6sin(atan(⅔)) units.
= 2(0.5*4.80740170062*6sin(atan(⅔)))
= 16 square units.
