Mathematics Question and Solution
Calculating OM/ON.
Let x be the radius of the inscribed circle.
a = (4+x) units.
a is AC.
b = (6+x) units.
b is AB.
c = 4+6
c = 10
c is BC.
Calculating x.
a²+b² = c²
(4+x)²+(6+x)² = 10²
16+8x+x²+36+12x+x² = 100
2x²+20x+52 = 100
2x²+20x-48 = 0
x²+10x-24 = 0
(x+5)² = 24+5²
(x+5)² = 49
x+5 = ±√(49)
x = -5±7
Therefore;
x ≠ -5-7
x = -5+7
x = 2 units.
Again, x is the radius of the inscribed circle.
Recall.
a = 4+x
And x = 2 units.
a = 4+2
a = 6 units.
Recall Again.
b = 6+x
And x = 2 units.
b = 6+2
b = 8 units.
It implies;
sind = a/c
d = asin(6/10)
d = asin(3/5)°
d is angle ABC.
Notice.
Length BC is parallel to length MN.
It implies;
d is also angle ANM.
tand = x/e
And x = 2 units.
tan(asin(3/5)) = 2/e
e = 8/3 units.
sind = x/f
And x = 2 units.
sin(asin(3/5)) = 2/f
3/5 = 2/f
3f = 10
f = 10/3 units.
f is ON.
g = x+e
g = 2+(8/3)
g = 14/3 units.
g is AN.
cosd = g/h
cos(asin(3/5)) = (14/3)/h
h = 35/6 units.
h is MN.
j = h-f
j = (35/6)-(10/3)
j = (35-20)/6
j = 15/6
j = 5/2 units.
j is OM.
It implies;
OM/ON is;
(5/2)÷(10/3)
= (5/2)*(3/10)
= ¾
