Value of angles of a quadrilateral

Question

$ABCD$ is a rectangle, $\overline{AC}$ and $\overline{BD}$ are its two diagonals, $O$ their intersection point, and $\angle COD=68^{\circ}$. What is the value of $\angle (BAO-OBC)$?

Answer 1

1. $\alpha = \angle COD = 68^{\circ}$ $\quad$ (is given)
2. $\angle COD = \angle AOB = \alpha$ $\quad$ (angles of the opposed vertex $O$ are the same)
3. $\angle OAB = \angle OBA := \beta$ $\quad$ (triangle $AOB$ is isosceles)
4. $180^{\circ} = \alpha + 2\beta$ $\quad$ (the sum of inner angles of a triangle is always $180^{\circ}$)
5. $\beta = 56^{\circ}$ $\quad$ (solve for $\beta$ eq. in step 3)
6. Let $\gamma = \angle OBC$, then $\beta + \gamma = 90^{\circ}$ $\quad$ (we are working with a rectangle)
7. $\gamma = 90^{\circ} - \beta = 90^{\circ} - 56^{\circ} = 34^{\circ}$ $\quad$ (solve for $\gamma$ in eq in step 6.)

The original question is $\angle BAO- \angle OBC$, that is: $\beta - \gamma = 56^{\circ} - 34^{\circ} = 22^{\circ}_{\blacktriangle}$.