In the given figure `DeltaABC` is equilateral on side AB produced. We choose a point such that A lies between P and B. We now denote 'a' as the length of sides of `DeltaABC`, `r_1` as the radius of incircle `DeltaPAC` and `r_2` as the ex-radius of `DeltaPBC` with respect to side BC. Then `r_1 + r_2` is equal to

Answer

`00000.50`

Explanation

`/_T_(1)O_(1)R=60^(@)` and `/_AO_(1)R=0^(@)`
similarly `/_BO_(2)S=30^(@)`
`T_(1)T_(2)=T_(1)A+AB+BT_(1)=RA+RB+SB`
`=r_(1) tan 30^(@)+a+r_(2)tan30^(@)=(r_(1)+r_(2))/(sqrt(3))+a`
and `T_(1)^(')T_(2)^(')=T_(1)^(')C+T_(2)^(')C=a-RA+a-BS=2a-(r_(1)+r_(2))/(sqrt(3))`
Since common external tangents are equal `T_(1)T_(2)=T_(1)^(')T_(2)^(')`
`implies r_(1)+r_(2)=(sqrt(3)a)/2`