Triangle $\triangle ABC$ has a right angle at $C$, $\angle A = 60^\circ$, and $AC=10$. Find the radius of the incircle of $\triangle ABC$

See the following image :

The center of the  incircle will be the intersection  of the angle bisectors shown

And the find the x coordinate of the center by solving these two equations :

y  = tan (135) [x -10sqrt(3)]       and  y  = tan(60) [x - 10sqrt (3)] + 10

Set these equations equal   and we have

tan (135) [x -10sqrt(3)]   = tan(60) [x - 10sqrt (3)] + 10       simplify

-1 [x - 10sqrt (3) ]   =  sqrt (3) [ x  - 10sqrt (3) ] + 10

10sqrt (3)  -  x   =  sqrt (3) x  - 30 + 10

x ( sqrt (3) + 1 )  =  10sqrt (3) + 20

x ( sqrt (3)  + 1)   =  10 [ sqrt (3)  + 2 ]

x =   10 [  sqrt (3)  + 2 ] / [ sqrt (3)  + 1 ]            multiply  top /bottom  by  sqrt (3) - 1     and we have

x  = 10 [  sqrt (3)  + 2 ] [ sqrt (3) - 1 ] /  [ 3 - 1]

x =  10  [  sqrt (3)  + 2 ] [ sqrt (3) - 1 ] /  [ 2]

x =  5 [ 3 + sqrt (3)  - 2]

x  =  5 [ 1 + sqrt (3) ]  =  5  + 5sqrt (3)

And the y coordinate of the center is :

And    y =   -1 [ 5 + 5sqrt (3) - 10sqrt (3) ]    =  -1 [  5 - 5sqrt (3) ]  =  5sqrt (3) - 5  =

5 [ sqrt (3)  - 1 ]       and this is the radius of the incircle