in the figure <ABD = 50°, <DBE = 20°, then find <ACB and <AFB
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Step-by-step explanation:
ACB=50
o
,∠ACB=∠CAD(∵BC∣∣AD)
In △AOD,∠DAO+∠ADO+∠AOD=180
o
50+∠ADB+90
o
=180
o
∠ADB=180
o
−140
o
=40
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