square `ABDC` is a cyclic quadrilateral.
`therefore angle BDC+BAC =180^@`
`rArr115^@+angle BAC =180^@`
`rArrangle ACB=180^@-115^@`
` 65^@`
In `Delta ABC, AC` is the diameter of the circle.
`thereforeangle ABC =90^@`
Now, `angle ACB=180^@-(angle ABC +angle BAC) `
`=180^@=(90^@+65^@)`
`rArrangle ACB =25^@` .