+9673 Obviously, \(m\angle \text{BOA} = m\angle\text{DOC}= 64^{\circ}\)
Considering the sum of interior angles of \(\triangle \text{BOA}\),
\(m\angle \text{BAO} + m\angle \text{BOA} + m\angle \text{ABO} = 180^{\circ}\\ x + 62^{\circ} + 64^{\circ} = 180^{\circ}\\ x = 54^{\circ}\)
Then, considering the sum of interior angles of \(\triangle \text{DOC}\),
\(m\angle \text{DCO} + m\angle \text{DOC} + m\angle \text{CDO} = 180^{\circ}\\ y+64^{\circ}+54^{\circ}= 180^{\circ}\\ y=62^{\circ}\)
Finally, \(x + y = 54^{\circ} + 62^{\circ} = 116^{\circ}\).
Therefore, option B is correct.
+9673 Obviously, \(m\angle \text{BOA} = m\angle\text{DOC}= 64^{\circ}\)
Considering the sum of interior angles of \(\triangle \text{BOA}\),
\(m\angle \text{BAO} + m\angle \text{BOA} + m\angle \text{ABO} = 180^{\circ}\\ x + 62^{\circ} + 64^{\circ} = 180^{\circ}\\ x = 54^{\circ}\)
Then, considering the sum of interior angles of \(\triangle \text{DOC}\),
\(m\angle \text{DCO} + m\angle \text{DOC} + m\angle \text{CDO} = 180^{\circ}\\ y+64^{\circ}+54^{\circ}= 180^{\circ}\\ y=62^{\circ}\)
Finally, \(x + y = 54^{\circ} + 62^{\circ} = 116^{\circ}\).
Therefore, option B is correct.
