in the given figure angle AOD =100° find angle BCD if AB is diameter and BC tangent
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simra32247:
no ur ans is incorrect..sorry
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OBC=90
DOB=180-100=80
OBD+DOB+DCB+ODC=360
90+80+DCB+ODC=360
170+DCB+ODC=360
DCB+ODC=360-170=190
In triangleAOD
AO=OD (radius)
OAD=ODA(isoceles triangle)
OAD+ODA+AOD=180
2OAD+100=180(OAD=ODA)
2OAD=180-100=80
OAD=40=ODA
ODC=180-40=140
OBC+BOD+ODC+BCD=360
90+80+140+BCD=360
170+140+BCD=360
310+BCD=360
BCD=360-310=50
answer is 50
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