AB parallel to CD A= 35 D= 75 find x
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∵AB∥CD and BD is transversal.
∴x=40°(Alternate angles)
Now in △ABD
75°+y+x=180°(Sum of interior angles of a triangle)
⇒y=180°−75°−40°=180°−115°=65°
Now in △BCD
40°+(y−30°)+z=180°(Sum of interior angles of a triangle)
∴z=180°−40°−35°=180°−75°=105
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AB= CD-DA
180=35-75
180=40
220 or -140
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