If `O` is the centre of a circle, find the value of `x` in each of the following figures `:`

Explanation

(i) `/_BDC = /_ BAC = 30^(@) [ Delta` in the same segment]
In `DeltaBCD`, the sum of the angles is `180^(@)`.
`:. /_ CBD + /_ BCD + /_ BDC = 180^(@)`
`implies 70^(@) + x^(@)+ 30^(@) = 180^(@)`
`implies x= 80^(@)`
(ii) `/_ APB = 90^(@) `[ angle in a semicircle ].
`/_ BAP + /_APB + /_ABP = 180^(@)`
`implies /_ BAP+90^(@)+35^(@)= 180^(@)`
`implies /_ BAP = ( 180^(@)- 125^(@)) = 55^(@)`
Now, `x^(@) /_ BQP = /_ BAP = 55^(@)` [ `Delta ` in the same segment].
Hence, `x = 55`.
(ii) OA = OB `implies /_ OAB implies OBA =25^(@)`
OC = OA `implies /_ OAC = /_ OCA = 35^(@)`
`:. /_ BAC = /_ OAB + /_ OAC = (25^(@) + 35^(@)) = 60^(@)`
Now, `x^(@) = /_ BOC = 2 /_ BAC = (2 xx 60^(@)) = 120^(@)`
`:. x= 120`
(iv) We know that the sum of all angles at a point is `360^(@)`.
`:. 90^(@)+120^(@) + /_ BOC = 360^(@) implies /_BOC = 150^(@)`
`/_ BOC = 2 /_BAC implies 2x = 150 implies x = 75`
(v) `/_ ODB implies /_OAC = 50^(@)` [ angles in the same segment]
OB= OD `implies /_ OBD = /_ODB=50-x^(@)`
`implies x=50`.
(vi) Reflex `/_AOC = (360^(@)-130^(@))= 230^(@)`
`:. /_ ABC = (1)/(2) xx "reflex " /_ AOC= ((1)/(2) xx230^(@)) = 115^(@) = x^(@)`
`:. x= 115`