MCQ
Diagonals of a Parallelogram $ABCD$ intersect at $O$. If $\angle\text{BOC}=90^\circ,\ \angle\text{BDC}=50^\circ$ then $\angle\text{OAB} $ is:
- A$10^\circ$
- B$50^\circ$
- ✓$40^\circ$
- D$90^\circ$
✓
Answer
Correct option: C.
$40^\circ$
$\angle\text{BOC} + \angle\text{COD} = 180^\circ$ (linear pair).
$\angle\text{COD} = 180^\circ - 90^\circ = 90^\circ$
In $\angle\text{DOC} + \angle\text{DCO} + \angle\text{ODC} = 180^\circ$ (angle sum property).
$90^\circ + \angle\text{DCO} + 50^\circ = 90^\circ$
$\angle\text{DCO} = 180^\circ - 140^\circ = 40^\circ$
$\angle\text{DCO} = \angle\text{OAB} = 40^\circ$ (alternate angles).
$\angle\text{COD} = 180^\circ - 90^\circ = 90^\circ$
In $\angle\text{DOC} + \angle\text{DCO} + \angle\text{ODC} = 180^\circ$ (angle sum property).
$90^\circ + \angle\text{DCO} + 50^\circ = 90^\circ$
$\angle\text{DCO} = 180^\circ - 140^\circ = 40^\circ$
$\angle\text{DCO} = \angle\text{OAB} = 40^\circ$ (alternate angles).
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