Correct option is B
Given:
Two circles intersect at points C and D.
AB is a common tangent to the two circles at points A and B.
∠ADB=68∘\angle ADB = 68^\circ∠ADB=680
Concept Used:
Alternate segment theorem:
By alternate segment theorem
∠BCE=∠BAC∠ACD=∠ABC Solution:
joining C and D point , now,
By Alternate Segment theorem,
∠BAC=∠CDAand∠ABC=∠BDC As we know, ∠ADB=680
∠ADC+∠BDC=∠ADB=68∘\angle ADC + \angle BDC = \angle ADB = 68^\circ∠ADC+∠BDC=∠ADB=68∘
Now, in
△ABC ∠BAC+∠ABC+∠ACB=180∘68∘+∠ACB=180∘∠ACB=180∘−68∘=112∘
Alternate Mehtod:
∠ADC+∠ACB=1800So,∠ACB=1800−680=1120
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