ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If angleDBC = 55° and angle BAC = 45°, find BCD.

  • -6

Given: ∠BAC =  and ∠DBC = 

∠DBC = ∠DAC =   (angles in the same segment made by chord CD)

∠A = ∠DAC + ∠BAC =  +  = 

Since opposite angles of a cyclic quadilateral are supplementary,

∠A + ∠BCD = 

∠BCD = 

If AB = BC,

∠ACB = ∠BAC = 

∠ECD = ∠BCD – ∠ACB = 

Hope it helps =)

Thumbs up please!! =D

  • -4
 
 
 

 Given : angle BAC = 45 degree and angle DBC = 55 degree

Now, angle CDB = angle BAC   (angles in the same segment are equal)
therefore, angle CDB = 45 degree.
In triangle BDC, 
angle DBC + angle CDB + angle DCB = 180 degree  (since angle sum prop)
55 degree + 45 degree + angle DCB = 180 degree
angle DCB + 100 degree = 180 degree
angle DCB = 180 degree - 100 degree
angle DCB = 80 degree
hope it helps :)
 
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