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In the given figure, AB∥CD∥EF,∠APO=35oAB \parallel CD \parallel EF,\angle APO = 35^oAB∥CD∥EF,∠APO=35o and ∠CQO=43o\angle CQO = 43^o∠CQO=43o​. The valu
Question

In the given figure, ABCDEF,APO=35oAB \parallel CD \parallel EF,\angle APO = 35^o and CQO=43o\angle CQO = 43^o​. The value of EOP+EOQ\angle EOP + \angle EOQ  is:

A.

282o282^o​​

B.

290o290^o​​

C.

264o264^o​​

D.

276o276^o​​

Correct option is A

Given:

AB∥CD

EF∥CD

∠APO = 35

∠CQO = 43

Concept Used:

Alternate interior angles: When two parallel lines are cut by a transversal, alternate interior angles are equal.

Sum of corresponding angle: The sum of the angles is 180 

Solution: 

APO+EPO=180°(corresponding angles) 35°+EPO=180° EPO=18035=145°\angle APO +\angle EPO =180\degree \quad \text{(corresponding angles)} \\ \ \\ 35\degree + \angle EPO = 180\degree \\ \ \\ \angle EPO = 180 - 35 = 145\degree  

Similarly;

CQO+EOQ=180°(corresponding angles) 43°+EOQ=180° EOQ=18043=137°\angle CQO +\angle EOQ =180\degree \quad \text{(corresponding angles)} \\ \ \\ 43\degree + \angle EOQ = 180\degree \\ \ \\ \angle EOQ = 180 - 43 = 137\degree  

Now, 

EOP+EOQ=145+137=282°\angle EOP + \angle EOQ = 145 + 137 = 282\degree ​​

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