The diagonals of a rectangle ABCD meet at O, If =44^ , find . - Vidyadip

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.

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