Question
The diagonals of a rectangle ABCD meet at O, If $\angle\text{BOC}=44^\circ,$ find $\angle\text{OAD}.$
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Answer
The rectangle ABCD is given as: 
We have,$\angle\text{BOC}+\angle\text{BOA}=180^\circ$ (Linear pair)
$44^\circ+\angle\text{BOA}=180^\circ$
$\angle\text{BOA}=180^\circ-44^\circ$
$\angle\text{BOA}=136^\circ$
Since, diagonals of a rectangle are equal and they bisect each other. Therefore, in$\triangle\text{OAB},$ we have
OA = OB (Angles opposite to equal sides are equal.) Therefore,$\angle1=\angle2$
Now,in $\triangle\text{OAB},$ we have$\angle\text{BOA}+\angle1+\angle2=180^\circ$
$\angle\text{BOA}+2\angle1=180^\circ$
$2\angle1=44^\circ$
$\angle1=22^\circ$
Since, each angle of a rectangle is a right angle. Therefore,$\angle\text{BAD}=90^\circ$
$\angle1+\angle3=90^\circ$
$22^\circ+\angle3=90^\circ$
$\angle3=68^\circ$
Thus, $\angle\text{OAD}=68^\circ$ Hence, the measure of $\angle\text{OAD}$ is 68º.$\angle\text{C}+\angle\text{D}=150^\circ$
Hence, the sum of $\angle\text{C}$ and $\angle\text{D}$ is 150º.
We have,$\angle\text{BOC}+\angle\text{BOA}=180^\circ$ (Linear pair)$44^\circ+\angle\text{BOA}=180^\circ$
$\angle\text{BOA}=180^\circ-44^\circ$
$\angle\text{BOA}=136^\circ$
Since, diagonals of a rectangle are equal and they bisect each other. Therefore, in$\triangle\text{OAB},$ we have
OA = OB (Angles opposite to equal sides are equal.) Therefore,$\angle1=\angle2$
Now,in $\triangle\text{OAB},$ we have$\angle\text{BOA}+\angle1+\angle2=180^\circ$
$\angle\text{BOA}+2\angle1=180^\circ$
$2\angle1=44^\circ$
$\angle1=22^\circ$
Since, each angle of a rectangle is a right angle. Therefore,$\angle\text{BAD}=90^\circ$
$\angle1+\angle3=90^\circ$
$22^\circ+\angle3=90^\circ$
$\angle3=68^\circ$
Thus, $\angle\text{OAD}=68^\circ$ Hence, the measure of $\angle\text{OAD}$ is 68º.$\angle\text{C}+\angle\text{D}=150^\circ$
Hence, the sum of $\angle\text{C}$ and $\angle\text{D}$ is 150º.
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