Question
ABCD is a rectangle with $\angle B A C=68^{\circ}$. Determine the value of $\angle D B C$.
✓
Answer
We have, $A B C D$ is the rectangle and let $O$ be the Intereection ooint of $\triangle C$ and $B D$, where $\angle B A C=68^{\circ}$
In a rectangle, diagonals are equal.
$\therefore \quad A C=B D \Rightarrow \frac{A C}{2}=\frac{B D}{2} \Rightarrow O C=B O\quad$ [ $\because$ in a rectangle, diagonals bisect each other]
$\Rightarrow \quad \angle C B O=\angle O C B=x\quad$ [$\because$ angles opposite to the equal sides in a triangle are equal]
Also, we have $A B \| C D$ and $A C$ is transversal.
$\therefore \quad \angle C A B=\angle A C D=68^{\circ} \quad$ [alternate angles]
and $\angle B C D=90^{\circ} \Rightarrow \angle D C O+\angle O C B=90^{\circ}$
$\Rightarrow \quad \angle D C A+\angle O C B=90^{\circ}$
$\Rightarrow \quad 68^{\circ}+x=90^{\circ} \quad\left[\because \angle D C A=68^{\circ}\right]$
$\Rightarrow \quad x=90^{\circ}-68^{\circ}=22^{\circ} \Rightarrow \angle O C B=22^{\circ}$
$\Rightarrow \quad \angle O B C=22^{\circ} \quad[\because \angle O C B=\angle O B C]$
In a rectangle, diagonals are equal.
$\therefore \quad A C=B D \Rightarrow \frac{A C}{2}=\frac{B D}{2} \Rightarrow O C=B O\quad$ [ $\because$ in a rectangle, diagonals bisect each other]
$\Rightarrow \quad \angle C B O=\angle O C B=x\quad$ [$\because$ angles opposite to the equal sides in a triangle are equal]
Also, we have $A B \| C D$ and $A C$ is transversal.
$\therefore \quad \angle C A B=\angle A C D=68^{\circ} \quad$ [alternate angles]
and $\angle B C D=90^{\circ} \Rightarrow \angle D C O+\angle O C B=90^{\circ}$
$\Rightarrow \quad \angle D C A+\angle O C B=90^{\circ}$
$\Rightarrow \quad 68^{\circ}+x=90^{\circ} \quad\left[\because \angle D C A=68^{\circ}\right]$
$\Rightarrow \quad x=90^{\circ}-68^{\circ}=22^{\circ} \Rightarrow \angle O C B=22^{\circ}$
$\Rightarrow \quad \angle O B C=22^{\circ} \quad[\because \angle O C B=\angle O B C]$
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