Mathematics Question and Solution
Calculating length OB.
a² = 6²+2²
a = √(40)
a = 2√(10) units.
a is AC.
tanb = 6/2
b = atan(3)°
√(40)² = 2√(50)²-2√(50)²cosc
40 = 100-100cosc
cosc = 6/10
c = acos(3/5)°
c is angle AOC.
d = ½(180-c)
d = ½(180-acos(3/5))
d = 63.4349488229°
d is angle OAC and angle OCA.
e = b-d
e = atan(3)-63.4349488229
e = 8.1301023542°
e is angle OCB.
Therefore, the required length, ? (OB) is;
Let it be f.
f² = √(50)²+2²-2*2√(50)cos8.1301023542
f = 5.0990195136 units.
Again, f is OB, the required length.
