In the given figure, O is the centre of a circle in which `/_ OAB = 20^(@)` and `/_ OCB = 55^(@)`. Find ` (i) /_ BOC , (ii) /_ AOC`.

Answer

`(i) /_ BOC = 70^(@)` ` (ii) /_ AOC = 70^(@)`

Explanation

`OB=OC implies /_OBC = /_ OCB = 55^(@)`
`:. /_ BOC = 180^(@) - ( /_ OCB + /_ OBC )`
`= [ 180^(@) - (55^(@) + 55^(@))] = 70^(@)`
`OA = OB implies /_ OBA = /_ OAB = 20^(@)`
`implies /_ OAB = 180^(@) - ( /_ OAB + /_ OBA )`
`implies /_ AOB = 180^(@) - (20^(@) + 20^(@)) = 140^(@)`
`:. /_ AOC = /_ AOB -/_ BOC = 140^(@) = 70^(@) = 70^(@)`