A circle is inscribed in a triangle ABC. It
touches the sides AB and AC at M and N
respectively. If O is the centre of the circle and
/A = 70°, then what is /MON equal to ?
Answers
Answer:
Step-by-step explanation:
Let I be the incenter of the in-circle.
∠B
1
A
1
C
1
=
2
1
(π−A)
A
1
=
3
π
−
2
1
(A−
3
π
)
A
2
=
3
π
−
2
1
(A
1
−
3
π
)
=
3
π
−
2
1
[
3
π
−
2
1
(A−
3
π
)−
3
π
]
=
3
π
+(
2
1
)
2
(A−
3
π
)
A
3
=
3
π
+(
2
1
)
3
(A−
3
π
)
So on,
A
n
=
3
π
+(
2
1
)
n
(A−
3
π
)
similarly, B
n
=
3
π
+(
2
1
)
n
(B−
3
π
)
and C
n
=
3
π
+(
2
1
)
n
(C−
3
π
)
when n→∞ then (
2
1
)
n
becomes zero or
A
n
=B
n
=C
n
=
3
π
.
Hence the △ will be equilateral
Solution:
We know that tangents from an external point are equal in length.
Given :
AB = 12 cm
BC= 8 cm
AC = 10 cm
Now let us take A as external point,thus
AD = AF
let AD = AF = x cm
So, DB = 12-x = BE [tangent from external point B]
CE = 8-12+x = -4+x = CF [tangent from external point C]
Since perimeter of ∆ABC does not change so
12+8+10= x+x+12-x+12-x-4+x-4+x
30=2x+16
2x=30-16
2x=14
x = 7 cm
So, AD = AE = 7 cm
BD=BE = 12-7=5 cm
CE=CF= -4+7=3 cm
Hope it helps you.
