The circle inscribed in ABC triangle is tangent to triangles

The circle inscribed in ABC triangle is tangent to triangle\'s sides in M (AB), N(BC), P(AC). PA=2,PC=3, C=pi/3, calculate the length of MN segment.

Solution

In triangle ABC angle C=90 deg.

CA and CB touch the circle at O and M

Therefore CP = CN and Angles CPN = CMN =90 as tngents make 90 with radius. And so, CPON is a square and angle PON =90 .

Given PC=3 . So,CN=3 and ON= OM = OP =3, being radius.

In triangle ACP, being rt angled at P, OP= =sqrt(2^2+3^2I=sqrt13.

So, angle AOP= cos inverse (3/sqrt13) =33.69006752/

Angle MOP=2*angle AOP =67.8013504

Therefore angle MON = 360-(Angle MOP+ anglePON)=202.619865

Therefore MN= sqrt{2*radius^2*(1-cos angle MON)}

=sqrt{2*(3^2)(1-cos 202.619865)}=5.883484054

Hope this helps.