In the adjoining figur, O is the centre of the circle and AB is a diamter. If AB and CD are mutually perpendicualr to each other and `angle ADC=50^(@)`, then the value of `angle CAD`

Explanation

The angles in cricle produced by the chord AC are `angle ADC and angle ABC`.
`:. angle ABC= angle ADC=50^(@)`[ by theorem-34]
Again, the central angle `=angleAOB` and angle in circle `=angleACB` both produced by the are `overset(frown)(ADB)`.
`:. angle AOB=2 angle ACB`[ by theorem-34]
`or, 180^(@)=2 angle ACB [ :. angle AOB` = straight angle `=180^(@)]`

or `angle ACB=(180^(@))/(2)=90^(@)`
Then `angle CAB=180^(@)-(angleACB+ angle ABC) [ :.` sum of three angles of a triangles is `180^(@)]`
`=180^(@)-(90^(@)+50^(@))=180^(@)-140^(@)=40^(@)`
Again, AB and CD are perpendicular to each other. LEt AB and CD intersect each other at E.
`:. angle EAD=180^(@)-(angle ADE+angle AED)`
`=180^(@)-(50^(@)+90^(@)) [ :. angle ADE= angle ADC=50^(@) and angle AED=90^(@)]`
`=180^(@)-140^(@)=40^(@)`
Now, `angle CAD=angle EAC+ angle EAD`
`=40^(@)+40^(@)=80^(@) [ :. angleEAC= angle CAB= 40^(@) and angle EAD=40^(@)]`
`:. angle CAD= 80^(@)`