MCQ
The diagonals of a parallelogram $ABCD$ intersect at $O.$ if $\angle\text{BOC}=90^\circ$ and $\angle\text{BDC}=50^\circ,$ then $\angle\text{AOB}=$
- ✓$40^\circ $
- B$50^\circ $
- C$10^\circ$
- D$90^\circ$
✓
Answer
Correct option: A.
$40^\circ $
In a parallelogram $ABCD,$
$\angle\text{OAB}=\angle\text{OCB}$
In $\triangle\text{OCB}$
$\angle\text{OCD}+\angle\text{COD}+\angle\text{ODC}=180^\circ$
$\angle\text{COD}=90^\circ$
$\angle\text{ODC}=50^\circ$ (given)
$\angle\text{OCD}=180^\circ-90^\circ-50^\circ=40^\circ$
$\Rightarrow\angle\text{OAB}=\angle\text{OCD}=40^\circ$
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