Question

In the figure, `bar(AB)" || "bar(CD) and bar(AD)" || "bar(BC)`. If `angleABC=110^(@), angleACD=30^(@)`, then find `angleADC`.

Answer 1

Correct Answer - `110^(@)`

Given `bar(AB)" || "bar(CD), angleABC=110^(@) and angleACD=30^(@)`
`rArr angleCAB=30^(@)`
`rArr angleCAB=30^(@)" (Alternate angles)"`
`therefore In DeltaABC, angle BCA=180^(@)-110^(@)-30^(@)=40^(@)`
Since `bar(AD)" || "bar(BC), angleDAC=angleACB=40^(@)`
(Alternate angles)
In `DeltaADC, angleADC=180^(@)-30^(@)-40^(@)=110^(@)`