110°
Now,
Using exterior angle property of a triangle in ∆BEF,
We have, ∠BEF + ∠EFB = ∠ABE
→ ∠BED + ∠EFB = ∠ABE (°•°∠BEF = ∠BED)
→ 20° + ∠EFB = 130°
→ ∠EFB = 130° - 20°
→ ∠EFB = 110°
AB || CD and ED is transversal, thus
→ ∠EDC = ∠EFA
(°•° Corresponding angles are equal)
→ ∠EDC = ∠EFB (°•°∠EFA = ∠EFB)
→ ∠EDC = 110°
Answer:
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Answers & Comments
Answer :
110°
Concept to be used :
Solution :
Now,
Using exterior angle property of a triangle in ∆BEF,
We have, ∠BEF + ∠EFB = ∠ABE
→ ∠BED + ∠EFB = ∠ABE (°•°∠BEF = ∠BED)
→ 20° + ∠EFB = 130°
→ ∠EFB = 130° - 20°
→ ∠EFB = 110°
Now,
AB || CD and ED is transversal, thus
→ ∠EDC = ∠EFA
(°•° Corresponding angles are equal)
→ ∠EDC = ∠EFB (°•°∠EFA = ∠EFB)
→ ∠EDC = 110°
Hence, ∠EDC = 110°
Answer: