The side AB of the cyclic quadrilateral ABCD is extended to X . If `angleXBC=82^(@)and angleADB=47^(@)` Find the value of `angleBAC`.

Explanation

Given that `angleXBC=82^(@)`
`thereforeangleABC=180^(@)-angleXBC=180^(@)-82^(@)=98^(@)`
Again , two angles in circle produced by the arc AB are `angleADBandangleACB, thereforeangleADB=angleACB,thereforeangleACB=47^(@)` ,
Now in `DeltaABC,angleBAC+angleACB+angleABC=180^(@)`
or , `angleBAC+47^(@)+98^(@)=180^(@)" or "angleBAC=180^(@)-145^(@)=35^(@)`
Hence `angleBAC=35^(@)`