Question

In the adjoining figure, AC is the diameter of the circle. If `angleBDC=115^(@)`, then find the value of `angle ACB`.

Answer 1

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^@` .