If O is the centre of the circle , find the value of x in eaxh of the following figures.

Explanation

(i) In `Delta ADC`, we have
`/_ CAD + /_ ADC + /_ ACD =180^(@)`
`implies 50^(@) + 90^(@) + /_ ACD = 180^(@)`
`implies /_ ACD = 40^(@)`
But, `/_ ACD =90^(@)` [angle in a semicircle] .
`:. /_ ACD + x^(@) = 90^(@)`
`implies 40^(@)+x^(@) = 90^(@)`
`implies x= 50`
(ii) Since AOB is a straight line , we have
`/_ AOC + /_ BOC = 180^(@)`
`implies 130^(@)+/_BOC =180^(@)`
`implies/_BOC = 180^(@)-130^(@)= 50^(@)`
Now, `/_ DBC =(1)/(2) /_BOC =25^(@) implies x = 25`
(iii) `/_BDC = /_ BAC = 40^(@)` [ `Delta ` in the same segment ].
`/_ BDC + /_ DBC + /_ BCD = 180^(@)`
`:. 40^(@) +80^(@) +x^(@) = 180^(@) implies x =60`
(iv) OA= OC `implies /_ OCA = /_ OAC =50^(@)`
Side CO of `Delta OCA` is produced to B.
`:. ` ext. `/_ AOB = /_OAC + /_ OCA = (50^(@)+ 50^(@)) = 100`
Now, `/_ BDA = (1)/(2) /_AOB = ((1)/(2) xx 100^(@))= 50^(@)`
Henc, `x = 50`
(v) In `Delta` OBA, we have OA =OB.
`:. /_ OBA= /_OAB = 50^(@) implies /_ CBA = 50^(@)`
Now, `/_ CDA = x^(@) = /_CBA = 50^(@) =50^(@) [ Delta ` in the same segment ].
`:. x= 50`.
(vi)`OA=OB` [ each equal to the radius of the circle ]
`implies /_ OBA = /_ OAB = x^(@)`
`:. x^(@) +x^(@) + 40^(@) = 180^(@) implies 2x^(@) = 140^(@) implies x= 70`.