In the adjoining firgure O is the centre of the centre and AB is the diameter. If `angle BCE=20^(@), angle CAE= 25^(@)`, then find the value of `angle AEC`.

Explanation

The angle in circle produced by the chrod CA are `angle AEC and angle ABC`.
`:. angleAEC= angle ABC.......(1)`
Again, the angles in circle produced by the chord BE are `angle BAE and angle BCE`.
`:. angle BAE= angle BCE=20^(@)[ :. angleBCE= 20^(@)]`
Similarly, the angels in circle produced by the chord CE are `angle CBE and angle CAE, :. angle CBE= angle CAE=25^(@) [ :. angleCAE=25^(@)]`
Again, `angle AEB` is a semicircular angle `:. angleAEB=90^(@)`
Then in `Delta ABE, angle ABE=180^(@)-(angle BAE+angleAEB)=180^(@)-(20^(@)+90^(@))=70^(@)`
`:. angle ABC= angleABE- angleBE=70^(@)-25^(@)=45^(@)`
`:.` from (1) we get, `angle AEC= angle ABC=45^(@) :. angleAEC =45^(@)`