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In the figure (not drawn to scale), ABC and DEF are two triangles, CA is parallel to FD and CFBE is a straight line. Find the value of x+y
  • a)
    185o
  • b)
    134o
  • c)
    148o
  • d)
    176o
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In the figure (not drawn to scale), ABC and DEF are two triangles, CA ...
∠FCA = ∠BFD  
(Corresponding angles) 
⇒ x = 51o 
Now, in ΔABC y = 51o+83o (Exterior angle property) 
⇒ y = 134So, x+y = 51o+134= 185o
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In the figure (not drawn to scale), ABC and DEF are two triangles, CA is parallel to FD and CFBE is a straight line. Find the value of x+ya)185ob)134oc)148od)176oCorrect answer is option 'A'. Can you explain this answer?
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