Question
If $ABCD$ is a rectangle with $\angle\text{BAC}=32^\circ,$ find the measure of $\angle\text{DBC}.$
✓
Answer
Figure is given as :
Suppose the diagonals $AC$ and $BD$ intersect at $O$.
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, $\angle\text{OAB}=\angle\text{OBA}$
$\angle\text{BAC}=\angle\text{DBA}$ But,
$\angle\text{BAC}=32^\circ$ $\angle\text{DBA}=32^\circ$
Now, $\angle\text{ABC}=90^\circ$
$\angle\text{DBA}+\angle\text{DBC}=90^\circ$
$32^\circ+\angle\text{DBC}=90^\circ$
$\angle\text{DBC}=58^\circ$
Hence, the measure of $\angle\text{DBC}$ is $58^\circ$.
Suppose the diagonals $AC$ and $BD$ intersect at $O$.
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, $\angle\text{OAB}=\angle\text{OBA}$
$\angle\text{BAC}=\angle\text{DBA}$ But,
$\angle\text{BAC}=32^\circ$ $\angle\text{DBA}=32^\circ$
Now, $\angle\text{ABC}=90^\circ$
$\angle\text{DBA}+\angle\text{DBC}=90^\circ$
$32^\circ+\angle\text{DBC}=90^\circ$
$\angle\text{DBC}=58^\circ$
Hence, the measure of $\angle\text{DBC}$ is $58^\circ$.
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