In the adjoining figure, O is the centre of the circle AC is a diameter and chord DE is parallel to the diameter AC. If `angle CBD=60^(@)` , find the value of `angle CDE`

Explanation

Let us join, DO . Then the central angle produced by the chord CD is `angle COD` and the angle in circle by the same chord CD is `angleCBD`.
`:. angleCOD=2 angle CBD`[ by theorem-34]
or, `angle COD=2xx60^(@), or angle COD=120^(@)`
`:. "in" Delta COD, angle OCD+ angle ODC=180^(@)- angle COD=180^(@)-120^(@)`
`[ :.` the sum of three angles of a triangle is `180^(@)`
`:. angle OCD+ angle OCD=60^(@)[ :. angle ODC= angle OCD]` ltvbrgt `or, 2 angle OCD=60^(@), or angle OCD=(60^(@))/(2)=30^(@)`
Again, `AC||DE, :. angleACD= angle CDE[ :.` altgernate angles]
or, `angle CDE= angle OCD [ :. angle ACD and angle OCD` are same angles]
Heneh the value of `angle CDE=30^(@)`