Answer :

Consider a Parallelogram ABCD,


Diagonals AC and BD intersect at O.


According to the problem it is given that DAC = 400, CAB = 350 and DOC = 1100.


Since AC is a straight line it’s angle AOC is 1800.


So, AOC can be written as


AOC = AOD + COD ...... (1)


AOC = AOB + BOC ...... (2)


Substituting the values in the eq(1) we get,


1800 = AOD + 1100


AOD = 1800 - 1100


AOD = 700 ...... (3)


Since BD is a straight line it’s angle BOD is 1800.


So, BOD can be written as


BOD = DOC + BOC ...... - (4)


Substituting the values in eq(4) we get,


1800 = BOC + 1100


BOC = 1800 - 1100


BOC = 700 ...... - (5)


Using eq. (2) and (5) we get,


1800 = AOB + 700


AOB = 1800 - 700


AOB = 1100 ...... - - (6)


From the alternative angles property of traversal line between two Parallel lines, we can write


OCD = OAB = 350


BCO = DAO = 400


We sum that sum of all angles in a triangle is 1800.


From the ΔAOD, ΔDOC, ΔCOB, ΔAOB we can write,


DAO + AOD + ODA = 1800


DOC + OCD + CDO = 1800


COB + OBC + BCO = 1800


AOB + OBA + BAO = 1800


On substituting the known values in these equations we get,


400 + 700 + ODA = 1800


1100 + ODA = 1800


ODA = 1800 - 1100


ODA = 700


1100 + 350 + CDO = 1800


1450 + CDO = 1800


CDO = 1800 - 1450


CDO = 350.


400 + OBC + 700 = 1800


OBC + 1100 = 1800


OBC = 1800 - 1100


OBC = 700.


1100 + OBA + 350 = 1800


OBA + 1450 = 1800


OBA = 1800 - 1450


OBA = 350


We know that CBO = CBD = 700 and ACB = OCB = 400


And also ADC = ADO + ODC


ADC = 700 + 350


ADC = 1050.


The values of ABO, ADC, ACB, CBD is 350, 1050, 400, 700.


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