Mathematics Question and Solution
6²+6²-2*6*6cosa = 12²+3²+2*12*3cosa
72-72cosa = 153+72cosa
144cosa = 81
a = acos(9/16)
a = 55.7711336722°
b = 180-a
b = 124.2288663278°
6²+12²-2*6*12cosc = 6²+3²+2*6*3cosc
180-144cosc = 45+36cosc
180cosc = 135
c = acos(3/4)
c = 41.4096221093°
d = 180-c
d = 138.5903778907°
e = 180-a-c
e = 82.8192442185°
(12/sin82.8192442185) = (f/sin55.7711336722)
f = 10 units.
g = 180-b-c
g = 14.3615115629°
(6/sin14.3615115629) = (h/sin124.2288663278)
h = 20 units.
Therefore, the required length, x is;
x² = 10²+20²-2*10*20cos41.4096221093
x = 14.1421356237 units.
