Mathematics Question and Solution
Calculating shaded area (triangle BCE).
Let a be the side length of the square.
5² = a²+3²-2*3*acosb
25 = a²+9-6acosb
16 = a²-6acosb --- (1).
8² = a²+3²-2*3*acos(90-b)
64 = a²+9-6a(cos90cosb+sin90sinb)
55 = a²-6asinb --- (2).
Dividing (2) by (1).
55/16 = 1+tanb
tanb = ((55/16)-1)
b = atan(39/16)
b = 67.6937949451°
b is angle CBE.
(8/sin67.6937949451) = (3/sinc)
c = 20.3001576321°
d = 180-b-c
d = 92.0060474228°
Therefore, shaded area (triangle BCE) is;
½*3*8sin92.0060474228
= 11.9926456548 square units.
= 12 square units.
