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in Mathematics by (106k points)

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 700 , ∠BAC is 300 , find ∠BCD. Further, if B = BC, find ∠ECD. 

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∠CDB = ∠BAC = 300 ...(i) [Angles in the same segment of a circle are equal] 

∠DBC = 700 ....(ii) 

 In ∆BCD, 

∠BCD + ∠DBC + ∠CDB = 1800 

[Sum of all he angles of a triangle is 1800

⇒ ∠BCD + 700 + 30.0 = 1800 [Using (i) and (ii)] 

⇒ ∠BCD + 1000 = 1800 

⇒ ∠BCD = 1800 - 100

⇒ ∠BCD = 800 ...(iii) 

 In ∆ABC, AB = BC 

∴ ∠BCA = ∠BAC = 300 ...(iv) 

[Angles opposite to equal sides of a triangle are equal] 

Now, ∠BCD = 800 [From (iii)] 

⇒ ∠BCA + ∠ECD = 800 

⇒ 300 + ∠ECD = 800 

⇒ ∠ECD = 800 - 300 

⇒ ∠ECD = 500

by (135 points)
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