In the adjoining figure, O is the center of the circle and AC is a diameter of it If DC||EB, `angleAOB=80^(@)andangleACE=10^(@)` , then find the value of `angleBED.

Explanation

Given that DC||EB and CE is their transversal.
`thereforeangleDCE=angleBEC[because"alternate angle"]` . . . . . . . .(1)
Now , `angleAOB=2angleACB[because` central angle is twice of angle of angle in circle]
Math (X)-20
or, `80^(@)=2angleACB[becauseangleAOB=80^(@)"given"]`
or ` angleACB=(80^(@))/(2)=40^(@)`
Also `angleBOC=180^(@)-angleAOB=180^(@)-80^(@)=100^(@)`
Again , the central angle produced by the chord `overset(frown)(BC)` is `angleBOC` and angle in circle is `angleBEC`
`thereforeangleBEC=(1)/(2)angleBOC=(1)/(2)xx100^(@)=50^(@)[becauseangleBOC=100^(@)]`
`thereforeangleECD=50^(@)` [by (1)]
Also , `angleBED+angleBCD=180^(@)[therefore` BCDE is a cyclic quadrilateral]
or, `angleBED=180^(@)-angleBCD=180^(@)-100^(@)=80^(@)`
`[becauseangleBCD=angleACB+angleACE+angleECD=40^(@)+10^(@)+50^(@)=100^(@)]thereforeangleBED=80^(@)`