Question
In the given figure, $ABCD$ is a rectangle in which diagonal $AC$ is produced to $E$. If $\angle\text{ECD}=146^\circ,$ find $\angle\text{AOB}.$
✓
Answer
$ABCD$ is a rectangle With diagonal $AC$ produced to point $E$.
We have $\angle1+\angle\text{DCE}=180^\circ$ (Linear pair)
$\angle1+146^\circ=180^\circ$
$\angle1=34^\circ$ We know that the diagonals of a parallelogram bisect each other.
Thus $OC = OD$ Also, angles opposite to equal sides are equal.
Therefore, $\angle\text{ODC}=34^\circ$ By angle sum property of a traingle
$\angle\text{ODC}+\angle1+\text{COD}=180^\circ$
$34^\circ+34^\circ+\text{COD}=180^\circ$
$68^\circ+\angle\text{COD}=180^\circ$
$\angle\text{COD}=112^\circ$
Also, $\angle\text{COD}$ and $\angle\text{AOB}$ are vertically opposite angles.
Therefore, $\angle\text{AOB}=112^\circ$
Hence, the required measure for $\angle\text{AOB}$ is $112^\circ$.
We have $\angle1+\angle\text{DCE}=180^\circ$ (Linear pair)
$\angle1+146^\circ=180^\circ$
$\angle1=34^\circ$ We know that the diagonals of a parallelogram bisect each other.
Thus $OC = OD$ Also, angles opposite to equal sides are equal.
Therefore, $\angle\text{ODC}=34^\circ$ By angle sum property of a traingle
$\angle\text{ODC}+\angle1+\text{COD}=180^\circ$
$34^\circ+34^\circ+\text{COD}=180^\circ$
$68^\circ+\angle\text{COD}=180^\circ$
$\angle\text{COD}=112^\circ$
Also, $\angle\text{COD}$ and $\angle\text{AOB}$ are vertically opposite angles.
Therefore, $\angle\text{AOB}=112^\circ$
Hence, the required measure for $\angle\text{AOB}$ is $112^\circ$.
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