In ` A B C ,\ /_B=/_C` and ray `A X` bisects the exterior angle `/_D A CdotIf\ /_D A X=70^0,\ t h e n\ /_A C B=` `35^0` (b) `90^0` (c) `70^0` (d) `55^0`

Explanation

Ray AX bisects exterior angle `/_DAC.
∴ /_CAD = 2 × /_DAX
/_CAD = 2 × 70
/_CAD = 140@`
By exterior angle theorem, If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles :
`Ext. /_CAD = /_B + /_C
140@ = /_C + /_C `
[Given :` /_B = /_C`]
`140@ = 2∠C
/_C = 140@/2
/_C = 70@
/_ACB = 70@`.