In a parallelogram ABCD, the diagonals bisect each other at O. If…

Question

In a parallelogram ABCD, the diagonals bisect each other at O. If $\angle\text{ABC}=30^\circ$, $\angle\text{BDC}=10^\circ$ and $\angle\text{CAB}=70^\circ$.
Find: $\angle\text{DAB},\angle\text{ADC},\angle\text{BCD},\angle\text{AOD},\angle\text{DOC},\angle\text{BOC}\\\angle\text{AOB},\angle\text{ACD},\angle\text{CAB},\angle\text{ADB},\angle\text{ACB},\angle\text{DBC}\\\text{and}\ \angle\text{DBA.}$

✓

Answer

In parallelogram ABCD, diagonal AC and ED bisect each other at O.
$\angle\text{ABC}=30^\circ$, $\angle\text{CAB}=70^\circ$ and $\angle\text{BDC}=10^\circ$
$\angle\text{ADC}=\angle\text{ABC}=30^\circ$ (Opposite angles)
and $\angle\text{ADB}=\angle\text{ADC}-\angle\text{BDC}$
$30^\circ-10^\circ=20^\circ$
AB || DC and AC is the transversal
$\angle\text{ACD}=\angle\text{CAB}=70^\circ$ (Alternate angles)
AB || DC and BD is transversal
$\angle\text{CDB}=\angle\text{ABD}=10^\circ$ (Altemate angles)
In $\triangle\text{ABC}$
$\angle\text{CAB}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ (Sum of anlges of a triangle)
$\Rightarrow70^\circ+30^\circ+\angle\text{BCA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{BCA}=180^\circ$
$\angle\text{BCA}=180^\circ-100^\circ$
$\angle\text{BCA}=80^\circ$
$\angle\text{BCD}=\angle\text{BCA}+\angle\text{ACD}=80^\circ+70^\circ$
$\angle\text{BCD}=150^\circ$
$\angle\text{BCD}=\angle\text{DAB}$ (Opposite angles)
$\angle\text{DAB}=150^\circ$ and
$\angle\text{CAD}=150^\circ-70^\circ$
$\angle\text{CAD}=80^\circ$
In $\triangle\text{OCD}$,
$\angle\text{ODC}+\angle\text{OCD}+\angle\text{COD}=180^\circ$ (Angles of a triangle)
$\Rightarrow\angle\text{ACD}+\angle\text{ACD}+\angle\text{COD}=180^\circ$
$\Rightarrow70^\circ+10^\circ+\angle\text{COD}=180^\circ$
$\Rightarrow80^\circ+\angle\text{COD}=180^\circ$
$\Rightarrow\angle\text{COD}=180^\circ-80^\circ$
$\angle\text{COD}=100^\circ$
But $\angle\text{AOD}+\angle\text{COD}=180^\circ$ (Linear pair)
$\Rightarrow\angle\text{AOD}+100^\circ=180^\circ$
$\Rightarrow\angle\text{AOD}=180^\circ-100^\circ$
$\Rightarrow\angle\text{AOD}=80^\circ$
But $\angle\text{AOB}=\angle\text{COD}$
and $\angle\text{BOC}=\angle\text{AOD}$ (Vertically opposite angles)
$\angle\text{AOB}=100^\circ$ and $\angle\text{BOC}=80^\circ$
Hence,
$\angle\text{DAB}=150^\circ$, $\angle\text{ADC}=30^\circ$, $\angle\text{BCD}=150^\circ$
$\angle\text{AOD}=80^\circ$, $\angle\text{DOC}=100^\circ$
$\angle\text{BOC}=80^\circ$, $\angle\text{AOB}=100^\circ$
$\angle\text{ACD}=70^\circ$, $\angle\text{CAB}=70^\circ$, $\angle\text{AD}=20^\circ$,
$\angle\text{ACB}=80^\circ$, $\angle\text{DBC}=20^\circ$, $\angle\text{DBA}=10^\circ$