Two circles with centres A and B inrtersect each other at the points C and D. The centre B of the other circle lies on the circle with centre A. if `angle CQD=70^(@)`, then find the value of `angle CPD`.

Explanation

Let us join `A,C, B,C, D and B,D`
Now, in the circle with centre at B, angle in circle produced by the are `overset(frown)(CD)= angle CQD` and central angle `=angle CBD`.
`:.` be theorem `-34 angle CBD= 2 angle CQD= 2xx70^(@) [ :. angleCQD= 70^(@)]= 140^(@)`,
Again in the circle with centre A, the central angle produced by the are `overset(frown)(CPD)`= reflex `angle CAD` and angle in circle `=angle CBD`
`:.` by theorem -34 reflex `angleCAD = 2 angle CBD`
` or, 360^(@)-angleCAD= 2 xx 140^(@)[ :. angle CBD= 140^(@)] or, angleCAD= 360^(@)-280^(@)=80^(@)`
Again, in the circle with centre A, the central angle produced by the are `overset (frown) (CD)= angleCAD` and its angle in circle `=angleCPD.`
By theorem `-34 angle CAD= 2 angle CPD`
or, `80^(@)=2 angle CPD[ :. angleCAD=80^(@)]`
or, `angle CPD=(80^(@))/(2)=40^(@)`
Hence the value of `angle CPD= 40^(@)`