Mathematics Question and Solution
a² = (2r)²+r²
a² = 5r²
a = √(5)r units.
a is BD.
Where r is the radius of the circle.
b = ½(a)
b = ½(√(5)r) units because BD, points B and D are symmetry with respect to length AC.
tanc = 2r/r
c = atan(2)°
c is angle DBF.
d = 90-c
d = (90-atan(2))°
tane = 2/1
e = atan(2)°
e is angle BAC.
tanf = 1/2
f = atan(½)°
f is angle ACB.
Notice!
c (angle DBF) + f (angle ACB) = 90°
Or
atan(2) + atan(½) = 90°
It implies;
(½(BD))BC is a right-angled triangle.
Notice!
½(BD) = ½(√(5)r) units.
Calculating ½(BD)C.
BC = 2 units.
Therefore;
cos(atan(½)) = g/2
g = 2cos(atan(½))
g = 1.788854382 units.
g is ½(BD)C.
Let ½(BD) = G
It implies;
g = CG.
Where (½(BD))BC is now GBC which is a right-angled triangle.
Therefore r, radius of the circle is;
Observing Pythagoras Rule.
(BC)² = (BG)²+(CG)²
2² = (½(√(5)r))²+1.788854382²
4-3.2 = ¼(5r²)
0.8*4 = 5r²
3.2 = 5r²
r² = 0.64
r² = (64/100)
r = 8/10
r = ⅘ units.
r = 0.8 units.
