The diagonals of a rectangle `A B C D` meet at `Odot` If `/_B O C=44^@,` find `/_O A Ddot`

Explanation

It is given that
`angle BOC+ angle BOA`=`180^@`
` 44^@+ angle BOA` = `180^@`
` angle BOA`=`180^@- 44^@`
`angle BOA`=`136^@`
Since diagonals of a rectangle are equal and they bisect each other .
Therfore, In ` triangle OAB`
` OA`=`OB`( angles opposite to equal sides are equal) and
`angle1`=`angle 2`
Now ,In `triangle OAB`
`angle 1 +angle 2+ angle BOA`=`180^@`
`2xxangle 1+136^@= 180^@`
`angle 1`=`22^@`
Since angle of rectangle is a right angle.
Therefore, `angle BAD`=`90^@`
` angle 1+ angle 3` =`90^@`
`angle 3`=`68^@`
So,`angleOAD`=`68^@`