Question 957815
If we take the bisector from each corner and find the intersection of the bisectors, that will be the center of the inscribed circle. 
The distance from the center of the circle to each side of the triangle will be equal to the incenter circle radius.
If you make the triangle PQR, then it's true that,
{{{PU+UR=PR}}}
{{{QT+TR=QR}}}
{{{PS+SQ=PQ}}}
I'll call {{{C}}} the center of the inscribed circle.
So then {{{PC}}},{{{QC}}},and {{{RC}}}, are the lengths of the bisectors from each vertex.
You can form right triangles
{{{PCS}}} with legs {{{PS}}}, {{{CS}}}, and hypotenuse {{{PC}}}.
{{{PCU}}} with legs {{{PU}}}, {{{CU}}}, and hypotenuse {{{PC}}}.
{{{QCS}}} with legs {{{QS}}}, {{{CS}}}, and hypotenuse {{{QC}}}.
{{{QCT}}} with legs {{{QT}}}, {{{CT}}}, and hypotenuse {{{QC}}}.
{{{RCU}}} with legs {{{RU}}}, {{{CU}}}, and hypotenuse {{{RC}}}.
{{{RCT}}} with legs {{{RT}}}, {{{CT}}}, and hypotenuse {{{RC}}}.
but
{{{CS=CU=CT=r}}}
So then since the triangles share two equal sides and an angle then,
{{{PS=PU=u}}}
{{{QS=QT=v}}}
{{{RU=RT=w}}}
Then substituting from above,
{{{PU+UR=PR}}}
1.{{{u+w=18}}}
.
.
{{{QT+TR=QR}}}
2.{{{v+w=16}}}
.
.
{{{PS+SQ=PQ}}}
3.{{{u+v=20}}}
Substracting 1 from 3,
{{{u+v-u-w=20-18}}}
4.{{{v-w=2}}}
Then adding 2 and 4,
{{{v+w+v-w=16+2}}}
{{{2v=18}}}
{{{v=9}}}
Then,
{{{u+9=20}}}
{{{u=11}}}
and finally,
{{{9-w=2}}}
{{{w=7}}}
So then,
{{{PS=PU=11}}}
{{{QS=QT=9}}}
{{{RU=RT=7}}}