3)The diagonals AC and BD of parallelogram ABCD intersect at point E. l( AC) = 7cm. Then l ( AE) =
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Given in the rectangle ABCD ∠DAC=32°
⇒∠AOB=70°
⇒∠A=90°
⇒∠DAC+∠BAC=90°
⇒∠BAC=90°−32°=58°
From △AOB
⇒∠AOB+∠OAB+∠OBA=180° [ Sum of angle property ]
⇒58
o
+70
o
+∠OBA=180°
∴∠OBA=180°−58°−70°=52°
We know that ∠B=90°
⇒∠ABO+∠DBC=90°
⇒∠DBC=90°−52°=38°
Hence, the answer is 38°.
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