6. In the given figure, AB || CD.
angle BAE= X angle AEC=20° angle DCE=130°.
Find the value of x.
Answers
Step-by-step explanation:
Expand CD toward AE intersecting at F such that CF || AB.
DCF is a straight line
so, 2 DCE + < FCE = 180°
130° + < FCE = 180° (given <DCE = 130°)
So <FCE = 180°-130° = 50°
Now angle FCE = 50 degree. Consider triangle FCE
<s (FCE + CEF + CFE) = 180°
50°+ 20° + <CFE = 180° (CEF = 20° given)
By solving this equation, we get angle CFE = 110 degree.
Now we have CF parallel to AB
So angle CFE = BAF { Corresponding angle } Hence, angle CFE BAF = 110°
i.e., <BAF = x = 110°
Answer:
Step by Step Explanation
From E, draw EF || AB || CD. EF || CD and CE is the transversal.
∴∠DCE+∠CEF=180
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[∵ Co-interior angles]
⇒x
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+∠CEF=180
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⇒∠CEF=(180−x
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)
Again, EF || AB and AE is the transversal.
∠BAE+∠AEF=180
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[∵ Co-interior angles]
⇒105
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+∠AEC+∠CEF=180
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⇒105
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+25
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+(180
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−x
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)=180
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⇒310−x
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=180
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Hence, x=130
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solution