Question

In a parallelogram ABCD, the diagonals bisect each other at O. If $\angle ABC={30}^{\circ },\angle BDC={10}^{\circ }and\angle CAB={70}^{\circ }$.

Find: $\angle DAB,\angle ADC,\angle BCD,\angle AOD,\angle DOC,\angle BOC,\angle AOB,\angle ACD,\angle CAB,\angle ADB,\angle ACB,\angle DBCand\angle DBA$.

Answer 1

In parallelogram ABCD, diagonal AC and BD bisect each other at O.

$\angle ABC={30}^{\circ },\angle CAB={70}^{\circ }and\angle BDC={10}^{\circ },$
$\angle ADC=\angle ABC={30}^{\circ }$, (Opposite angles)
and $\angle ADB=\angle ADC-\angle BDC={30}^{\circ }-{10}^{\circ }={20}^{\circ }$
$\because$ AB||DC and AC is the transversal
$\therefore \angle ACD=\angle CAB={70}^{\circ }$ (Alternate angles)
AB ||DC and BD is transversal
$\therefore \angle CDB=\angle ABD={10}^{\circ }$ (Alternate angles)
In $\Delta ABC$
$\angle CAB+\angle ABC+\angle BCA={180}^{\circ }$
(Sum of angles of a triangle)
$\Rightarrow {70}^{\circ }+{30}^{\circ }+\angle BCA={180}^{\circ }\Rightarrow {100}^{\circ }+\angle BCA={180}^{\circ }\angle BCA={180}^{\circ }-{100}^{\circ }={80}^{\circ }\therefore \angle BCD=\angle BCA+\angle ACD={80}^{\circ }+{70}^{\circ }={150}^{\circ }$
$\therefore \angle BCD=\angle DAB$ (Opposite angles)
$\Rightarrow \angle DAB={150}^{\circ }$
and $\angle CAD={150}^{\circ }-{70}^{\circ }={80}^{\circ }$
(Angles of a triangle)
$\Rightarrow \angle ACD+\angle ACD+\angle COD={180}^{\circ }\Rightarrow {70}^{\circ }+{10}^{\circ }+\angle COD={180}^{\circ }{80}^{\circ }+\angle COD={180}^{\circ }\Rightarrow \angle COD={180}^{\circ }-{80}^{\circ }={100}^{\circ }\therefore \angle COD={100}^{\circ }$
But $\angle AOD+\angle COD={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle AOD+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \angle AOD={180}^{\circ }-{100}^{\circ }={80}^{\circ }\Rightarrow \angle AOD={80}^{\circ }$
But $\angle AOB=\angle COD$ and $\angle BOC=\angle AOD$
(Vertically opposite angles)
$\therefore \angle AOB={100}^{\circ }and\angle BOC={80}^{\circ }$
Hence $\angle DAB={150}^{\circ },\angle ADC={30}^{\circ },\angle BCD={150}^{\circ }\angle AOD={80}^{\circ }\angle DOC={100}^{\circ }\angle BOC={80}^{\circ }\angle AOB={100}^{\circ }\angle ACD={70}^{\circ },\angle CAB={70}^{\circ },\angle ADB={20}^{\circ },\angle ACB={80}^{\circ }\angle DBC={20}^{\circ },\angle DBA={10}^{\circ }$