In the figure, `bar(DC)" || "bar(AB)`. If `angleACB=40^(@)` and `angleCAD=30^(@)`, AC is the bisector of `angleDAB`, then find the angle of `angleADC`.

Answer

`120^(@)`

Explanation

Given `angleCAD=30^(@), angleACB=40^(@)`
As AC is the bisector of `angleBAD, angle CAD=angleBAC=30^(@)`.
Given, `bar(AB)" || "bar(DC)`.
`therefore angleACD=angleBAC=30^(@)`
In `DeltaADC, angleADC=180^(@)-30^(@)-30^(@)=120^(@)`