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In ∆ABC
Angle ABC=40°
Angle CAB=60° (alternate angle of EDC)
so
/_ABC+/_CAB+/_ACB=180°
60°+40°+/_ACB=180°
◆/_ACB=80°◆
Since,AD is straight line
so,
/_ACB+x=180°
80°+x=180°
●x=100°●
Hope it will help you!
Angle ABC=40°
Angle CAB=60° (alternate angle of EDC)
so
/_ABC+/_CAB+/_ACB=180°
60°+40°+/_ACB=180°
◆/_ACB=80°◆
Since,AD is straight line
so,
/_ACB+x=180°
80°+x=180°
●x=100°●
Hope it will help you!
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