ABCD is a parallelogram if ∆ bac= 45° & ∆ ACB=50° find ∆ dab ,∆ ACD &∆ ADC
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ACB=50o,∠ACB=∠CAD(∵BC∣∣AD)
In △AOD,∠DAO+∠ADO+∠AOD=180o
50+∠ADB+90o=180o
∠ADB=180o−140o=40
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