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# help math

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143
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http://prntscr.com/mxbt4s

Mar 13, 2019

#1
+7709
+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
+7709
+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.

MaxWong Mar 13, 2019