Diagonals AC and BD of parallelogram ABCD intersect at O. If angleBOC = 90° and angle BDC = 50°, find angleOAB
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In the given details angle BOC= 90° and angle BDC = 50° Now angle BDC =DBA (ALTERNATE INTERIOR ANGLES ) AND ANGLE BOC= BOA (BY TAKING THE LINEAR PAIR ) THEN ,ANGLE AOB +OBA + OAB =180° 90°+50°+angle OAB =180°. 140°+ ANGLE OAB = 180° ANGLE OAB = 180° - 140°. ANGLE OAB = 40°. GOT IT
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40 degrees 4th option
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