Mathematics Question and Solution
tana = 3/4
a = atan(¾)°
a is angle ACB.
Let b be the radius of the three inscribed congruent circles.
c = (4-5b) units.
Calculating b.
tan(0.5atan(¾)) = b/(4-5b)
⅓ = b/(4-5b)
3b = 4-5b
8b = 4
b = ½ units.
b = 0.5 units.
Again, b is the radius of the three inscribed congruent circles.
Calculating r, radius of the bigger inscribed circle.
d = (r+0.5) units.
e = 3(½)-r
e = (1.5-r) units.
tanf = 4/3
f = atan(4/3)°
tan(0.5atan(4/3)) = r/g
½ = r/g
g = 2r units.
h = 3-g-0.5
h = (2.5-2r)
It implies;
(r+0.5)² = (1.5-r)²+(2.5-2r)²
r²+r+¼ = ¼(9)-3r+r²+¼(25)-10r+4r²
r²+r+¼ = ¼(34)-13r+5r²
4r²+4r+1 = 34-52r+20r²
16r²-56r+33 = 0
Resolving the above quadratic equation via completing the square approach to get r, radius of the bigger inscribed circle.
16r²-56r+33 = 0
r²-(7/2)r = (-33/16)
(r-(7/4))² = (-33/16)+(-7/4)²
(r-(7/4))² = (-33/16)+(49/16)
(r-(7/4))² = 1
r = (7/4)±1
It implies;
r = (7/4)-1
r = ¾ units.
r = 0.75 units.
Again, r is the radius of the bigger inscribed circle.
