In the given figure ab parallel
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CD is parallel to AB
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as we know that AB||CD therefore angle ABA = angle BCD= x° (alternate angle) now BC||CD therefore angle BCD+angleEDC=180°(because if the lines are parallel then the sum of the interior angles on the same side of transversal are supplementary) angle BCD +angle EDC=180° x°+75°=180° x=180°-75°= 105° so x=105°
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