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The diagonals of a parallelogram A B C...

The diagonals of a parallelogram `A B C D` intersect at `O` . If `/_B O C=90^@\ a n d\ /_B D C=50^@,` then `/_O A B=` `40^@` (b) `50^@` (c) `10^@` (d) `90^0`

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Given :The diagonals of a parallelogram ABCD intersect at O.
`angleBOC=90^@` and `angleBDC=50^@`
Solution : `therefore` `angleBOC=90^@`
`angleCOD=180^@-90^@`
`angleCOD=90^@`
In `triangle` COD
`angleOCD=90^@-50^@`
`angleOAB=angleOCD` ( Alternate angles )
...
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