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  • The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If DAC=32 and AOB=70, then DBC is equal to (A) 24º (B) 86º (C) 38º (D) 32º

The diagonals AC and BD of a parallelogram ABCD intersect each other at thepoint O. If \angle DAC=32^{\circ} and \angle AOB=70^{\circ}, then \angle DBCis equal to
(A) 24^{\circ}         (B) 86^{\circ}           (C) 38^{\circ}          (D) 32^{\circ}

Answers (1)

Answer: [C] 38^{\circ}
 

Solution.
Given \angle AOB=70^{\circ} and \angle DAC=32^{\circ}


\angle ACB=32^{\circ}       {AD\parallel BC and AC is a transversal line}
\angle AOB+\angle BOC=180^{\circ}
70^{\circ}+\angle BOC=180^{\circ}
\angle BOC=180^{\circ}-70^{\circ}=110^{\circ}
In \triangle BOC, we have
\angle BOC+\angle OCB+\angle CBO=180^{\circ}
110^{\circ}+32^{\circ}+\angle CBO=180^{\circ}
\angle CBO=180^{\circ}-110^{\circ}-32^{\circ}
\angle CBO=38^{\circ}
\angle DBC=\angle CBD=38^{\circ}
Hence option C is correct

Posted by

infoexpert26

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