Question
In a parallelogram ABCD, if $\angle\text{DAB}=75^\circ$ and $\angle\text{DBC}=60^\circ,$ then $\angle\text{BDC}=$
- 75°
- 60°
- 45°
- 55°
Solution:

In parallelogram ABCD,
$\angle\text{A}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-75^\circ=105^\circ$
$\angle\text{ADB}=\angle\text{DBC}$ (Alternate angles)
$\Rightarrow\angle\text{ADB}=60^\circ$
$\angle\text{BDC}=\angle\text{ADC}-\angle\text{ADB}=105^\circ-60^\circ=45^\circ$
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