Mathematics Question and Solution
a² = 3²+5²-2*3*5cos60
a = √(19) units.
a = 4.3588989435 units.
a is AC.
b² = 2²+1²-2*2*1cos60
b = √(3) units.
b = 1.7320508076 units.
b is CD.
(√(19)/sin120) = (√(3)/sinc)
c = 20.1283062882°
c is angle CAD.
d = 60-c
d = 39.8716937118°
d is angle ACD.
It implies, Area ABCD (area inscribed yellow cyclic quadrilateral) is;
Area triangle with height 5 units and base 3sin60 units + Area triangle with height √(19) units and base √(3)sin39.8716937118 units.
= 0.5*5*3sin60+0.5*√(19)*√(3)sin39.8716937118
= 6.4951905284+2.4199882366
= 8.915178765 square units.
