In the given figure, `O` is the centre of a circle and `/_ADC = 130^(@)`. If `/_ BAC = x^(@), ` then find the value of `x`.

Explanation

Since `ABCD` is a cyclic quadrilateral , we have
`/_ ABC + /_ ADC = 180^(@)`
`implies /_ ABC + 130^(@) = 180^(@)`
`implies /_ABC = 180^(@) -130^(@) = 50^(@)`
Also, `/_ ACB= 90^(@)` [angle in a semicircle] .
We know that the sum of the angles of a triangle si `180^(@)`.
`:. ` in `Delta ABC,` we have `/_ BAC + /_ ACB + /_ ABC = 180^(@)`
`implies x+ 90 + 50 = 180 implies x = 180 - 140 = 40`
Hence, `x= 40`