MCQ
In a parallelogram $ABCD$, if $\angle\text{DAB}=75^\circ$ and $\angle\text{DBC}=60^\circ,$ then $\angle\text{BDC}=$
- A$75^\circ$
- B$60^\circ$
- ✓$45^\circ$
- D$55^\circ$

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)
$\Rightarrow\angle\text{ADB}=60^\circ$
$\angle\text{BDC}=\angle\text{ADC}-\angle\text{ADB}=105^\circ-60^\circ=45^\circ$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
