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 Mar 13, 2019

Best Answer 

 #1
avatar+7458 
+1

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.

 Mar 13, 2019
 #1
avatar+7458 
+1
Best Answer

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.

MaxWong Mar 13, 2019

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