Question

In a parallelogram ABCD, the diagonals bisect each other at O. If ∠ABC = 30°, ∠BDC = 10° and ∠CAB = 70°. Find:
∠DAB, ∠ADC, ∠BCD, ∠AOD, ∠DOC, ∠BOC, ∠AOB, ∠ACD, ∠CAB, ∠ADB, ∠ACB, ∠DBC and ∠DBA.

Answer 1

\[\angle ABC = 30°\]

\[ \therefore \angle ADC = 30° \left( \text{ opposite angle of the parallelogram }\right)\]

\[\text{ and } \angle BDA = \angle ADC - \angle BDC = 30° - 10° = 20°\]

\[\angle BAC = \angle ACD = 70°(\text{ alternate angle })\]

\[\text{ In } \bigtriangleup ABC: \]

\[\angle CAB + \angle ABC + \angle BCA = 180°\]

\[70° + 30° + \angle BCA = 180°\]

\[ \therefore \angle BCA = 80°\]

\[\angle DAB = \angle DAC + \angle CAB = 70° + 80°= 150°\]

\[\angle BCD = 150° \left( \text{ opposite angle of the parallelogram } \right)\]

\[\angle DCA = \angle CAB = 70°\]

\[\text{ In } \bigtriangleup DOC: \]

\[\angle ODC + \angle DOC + \angle OCD = 180\]

\[10° + 70°+ \angle DOC = 180°\]

\[ \therefore \angle DOC = 100°\]

\[\angle DOC + \angle BOC = 180°\]

\[\angle BOC = 180° - 100°\]

\[\angle BOC = 80°\]

\[\angle AOD = \angle BOC = 80° \left( \text{ vertically opposite angles } \right) \]

\[\angle AOB = \angle DOC = 100° \left( \text{ vertically opposite angles } \right) \]

\[\angle CAB = 70° \left( \text{ given } \right)\]

\[\angle ADB = 20°\]

\[\angle DBA = \angle BDC = 10°(\text{ alternate angle })\]

\[\angle ADB = \angle DBC = 20°(\text{ alternate angle })\]