In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at O; and ∠DAC = 40° , ∠CAB = 35°; and ∠DOC = 110°. Calculate the measure of ∠ABO, ∠ADC, ∠ACB, and ∠CBD.

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.