Answer :

Consider a Parallelogram **ABCD**,

Diagonals **AC** and **BD** intersect at **O**.

According to the problem it is given that **∠****DAC = 40 ^{0},**

**∠**

**CAB = 35**and

^{0}**∠**

**DOC = 110**.

^{0}Since **AC** is a straight line it’s angle **∠****AOC** is **180 ^{0}**.

So, **∠****AOC** can be written as

⇒ **∠****AOC =** **∠****AOD +** **∠****COD** ...... (1)

⇒ **∠****AOC =** **∠****AOB +** **∠****BOC** ...... (2)

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

⇒ 180^{0} = ∠AOD + 110^{0}

⇒ ∠AOD = 180^{0} - 110^{0}

⇒ **∠****AOD = 70 ^{0}** ...... (3)

Since **BD** is a straight line it’s angle **∠****BOD** is **180 ^{0}**.

So, **∠****BOD** can be written as

⇒ **∠****BOD =** **∠****DOC +** **∠****BOC** ...... - (4)

Substituting the values in eq(4) we get,

⇒ 180^{0} = ∠BOC + 110^{0}

⇒ ∠BOC = 180^{0} - 110^{0}

⇒ **∠****BOC = 70 ^{0}** ...... - (5)

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

⇒ 180^{0} = ∠AOB + 70^{0}

⇒ ∠AOB = 180^{0} - 70^{0}

⇒ **∠****AOB = 110 ^{0}** ...... - - (6)

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

**∠****OCD =** **∠****OAB = 35 ^{0}**

**∠****BCO =** **∠****DAO = 40 ^{0}**

We sum that sum of all angles in a triangle is 180^{0}.

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

⇒ **∠****DAO +** **∠****AOD +** **∠****ODA = 180 ^{0}**

⇒ **∠****DOC +** **∠****OCD +** **∠****CDO = 180 ^{0}**

⇒ **∠****COB +** **∠****OBC +** **∠****BCO = 180 ^{0}**

⇒ **∠****AOB +** **∠****OBA +** **∠****BAO = 180 ^{0}**

On substituting the known values in these equations we get,

⇒ 40^{0} + 70^{0} + ∠ODA = 180^{0}

⇒ 110^{0} + ∠ODA = 180^{0}

⇒ ∠ODA = 180^{0} - 110^{0}

⇒ **∠****ODA = 70 ^{0}**

⇒ 110^{0} + 35^{0} + ∠CDO = 180^{0}

⇒ 145^{0} + ∠CDO = 180^{0}

⇒ ∠CDO = 180^{0} - 145^{0}

⇒ **∠****CDO = 35 ^{0}**.

⇒ 40^{0} + ∠OBC + 70^{0} = 180^{0}

⇒ ∠OBC + 110^{0} = 180^{0}

⇒ ∠OBC = 180^{0} - 110^{0}

⇒ **∠****OBC = 70 ^{0}**.

⇒ 110^{0} + ∠OBA + 35^{0} = 180^{0}

⇒ ∠OBA + 145^{0} = 180^{0}

⇒ ∠OBA = 180^{0} - 145^{0}

⇒ **∠****OBA = 35 ^{0}**

We know that **∠****CBO =** **∠****CBD = 70 ^{0}** and

**∠**

**ACB =**

**∠**

**OCB = 40**

^{0}And also **∠****ADC =** **∠****ADO +** **∠****ODC**

⇒ ∠ADC = 70^{0} + 35^{0}

⇒ **∠****ADC = 105 ^{0}**.

The values of **∠****ABO,** **∠****ADC,** **∠****ACB,** **∠****CBD** is **35 ^{0}, 105^{0}, 40^{0}, 70^{0}**.

Rate this question :

A field is in theKarnataka Board - Mathematics Part II

In a parallKarnataka Board - Mathematics Part II

In the figuKarnataka Board - Mathematics Part II