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Given, a trapezium ABCD in which ∠A=(x−20)∘,∠D=(x+40)∘
Since, in a trapezium, the angles between the parallel lines are supplementary,
(x−20)+(x+40)=180∘⇒x−20+x+40=180∘⇒2x+20∘=180∘⇒2x=(180∘−20∘)=160∘⇒x=80∘
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:
Given, a trapezium ABCD in which ∠A=(x−20)∘,∠D=(x+40)∘
Since, in a trapezium, the angles between the parallel lines are supplementary,
(x−20)+(x+40)=180∘⇒x−20+x+40=180∘⇒2x+20∘=180∘⇒2x=(180∘−20∘)=160∘⇒x=80∘
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