# In the figu

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

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

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