In the given figure, `AB` is a diameter of a circle with centre `O` and `CD "||" BA`. If `/_ BAC = 20^(@)` , find
(i) `/_ BOC ` (ii) `/_ COD `
(iii) `/_CAD` and (iv)` /_ ADC`

Explanation

(i) Arc `BC` subtends `/_ BOC ` at the centre an `/_ BAC` at a point on the remaining part of the circumstances
`:. /_ OC = 2 xx /_ ABC = 2 xx20^(@)= 40^(@)`
(ii) `CD "||" BA` and OC cuts them.
`:. /_ OCD = /_ BOC = 40^(@)` [alt. int. `/_ s` ] .
`:. /_ ODC = /_ OCD = 40^(@)` `[: ' /_ OC = OD = `radius ].
Now, in `Delta OCD `, we have
`/_ COD + /_ OCD + /_ODC = 180^(@)`
`implies /_ COD + 40^(@)+40^(@)=180^(@)" "[:' /_ OCD = /_ ODC = 40^(@) ]`
`implies /_ COD = 180^(@) - 80^(@) = 100^(@)`
(iii) Since the angle at the centre is double the angle at a point on the remaining part of the circumference, we have
`/_ CAD = (1)/(2) /_ COD = ((1)/(2) xx 100^(@))= 50^(@)`
(iv) Since ` BA ||CD` and `AD` cuts them, we have
`/_ ACD = /_ CAB = 20^(@) ` [alt. int. `/_s].`
In `Delta ACD`, we have
`/_ CAD + /_ ACD + /_ADC= 180^(@)`
`implies 50^(@) + 20^(@) + /_ ADC = 180^(@)`
`implies /_ ADC = 180^(@) -70^(@) = 110^(@)`