In the given figure, AB = BC, `angleBAC=70^(@)` and BC is produced to E, AC = CD and `angleADE` is the exterior angle of `DeltaADC`. Find the angle of `angleADE`.

Answer

`145^(@)`

Explanation

Since AB = BC, `angleBAC=angleACB=70^(@)`
`therefore angleABC=180^(@)-70^(@)-70^(@)=40^(@)`
`rArr angleACD=180^(@)-70^(@)=110^(@)`
AC = CD (given)
`rArr angleCAD=angleADC=(180^(@)-110^(@))/(2)=35^(@)`
`therefore angleADE=180^(@)-35^(@)=145^(@)`