In the isosceles `Delta ABC,AB=AC`. The circumcentre of `Delta ABC` is the centre O lies on the opposite side of BC in which a lies. If `Delta BOC=100^(@)`, then find the value of `angle ABC angle ABO`.

Explanation

In `Delta BOC, OB, OC [ :.` radii of same circle]
`:. angle OBC= angleOCB`
or, ` angle OBC+ angleOBC= angle OCB+ angleOBC`
or, `2 angleOBC= 180^(@)- angle BOC or, 2 angleOBC=180^(@) -100^(@)`
or, `2 angle OBC=810^(@) or, angleOBC=(80^(@))/(2)=40^(@)`

Again, the centralangle produced by the are `overset (frown) (BC)= angleBOC` and the angle in cricle `=angle BAC`.
By theorem -34, `angleBOC= 2 angle BAC`.
or, `angle BAC=(1)/(2) angle BOC=(1)/(2) xx 100^(@)=50^(@)`
Now, angle`ABC=180^(@)- (angleACB+ angle BAC)`
`=180^(@)- angle ABC-50^(@)[ :. angle ACB= angle ABC and angle BAC=50^(@)]`
`=130^(@)- angle ABC`
or, `2angle ABC= 130^(@), or angle ABC=(130^(@))/(2)=65^(@)`
Then `angleABO= angle ABC-angle OBC=65^(@)- 40^(@)=25^(@)`
Hence, `angle ABC=65^(@) and angle ABO=25^(@)`