In this chapter, we provide KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 pdf, free KSEEB solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 book pdf download. Now you will get step by step solution to each question.

## Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1

Question 1.

In Fig 3.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer:

∠AOC = ∠BOD = 40°

∠BOD = 40° (Date)

∠AOC = ∠BOD = 40° because vertifically opposite angles.

∴ ∠AOC + ∠BOE = 70°

40° + ∠BOE = 70°

∴ ∠BOE = 70° – 40° = 30°, and

∠AOD = 180° – ∠BOD = 180° – 40° = 140°

Reflex angle COE = ∠COA + ∠AOD + ∠BOD + ∠BOE

= 40° + 140° + 40° + 30°

∴ Reflex angle, ∠COE = 250°.

Question 2.

In Fig. 3.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer:

∠XOP + ∠POY = 180°

∵ straight supplementary

∠XOP + 90° =180°

∴ ∠XOP = 180° – 90° = 90°

But, ∠XOP = a + b = 90°

a : b = 2 : 3

2 + 3 = 5 Ratio

Ratio 5 means 90°

(=frac{2 times 90}{5}=2 times 18=36^{circ})

∴ If a = 36° then, ∠b = 54°.

∠XOM = ∠YON = b = 54° (∵ Vertically opposite angles)

∠XON + ∠YON = 180° (∵ Straight angle)

∴ c + 54° = 180°

c = 180 – 54

∴ c = 126°.

Question 3.

In Fig. 3.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer:

Data: In this figure, ∠PQR = ∠PRQ.

To prove: ∠PQS = ∠PRT

Proof: In ∆PQR, ∠Q = ∠R.

∴ This is an isosceles triangle.

Let ∠Q = 70°, then ∠R = 70° .

∠PQS + ∠PQR = 180°

∠PQS + 70° = 180°

∴ ∠PQS =180 – 70

∠PQS = 110° … (i)

Similarly,

∠PRT + ∠PRQ = 180°

∠PRT + 70° = 180°

∴ ∠PRT = 180 – 70

∠PRT =100 … (ii)

from (i) and (ii)

∠PQS = ∠PRT =110°

∴ ∠PQS = ∠PRS proved.

Question 4.

If Fig. 3.16, if x + y = w + z, then prove that AOB is a line.

Answer:

Data: In this fiure, ∠BOC = x°

∠AOC = y°

∠BOD = w°

∠AOD = z° and

x + y = w + z.

To Prove: AOB is a straight line.

Proof: x + y + w + z = 360° (∵ one completre angle)

But, x + y = w + z.

∴ x + y = w + z= (frac{360}{2}) = 180°

∴ x + y = 180°

∴ w + z = 180° proved.

But, ∠AOC and ∠BOC are adjacent angles,

∠AOC + ∠BOC = 180°

x = y = 180°

∴ ∠AOB = 180°.

∴ AOB is a straight line.

Question 5.

In Fig. 3.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

∠ROS = (frac{1}{2}) (∠QOS – ∠POS).

Answer:

Data: POQ is a straight line. Ray OR is perpendicular on straight line PQ. OS Ray is in between Rays OP and OR.

To Prove: ∠ROS = (frac{1}{2}) (∠QOS – ∠POS)

Proof : ∠QOR = ∠POR = 90° (Data)

∠POR = ∠POS + ∠ROS

∴ ∠ROS = ∠POR – ∠POS

∠ROS = 90°- ∠POS (i)

Now, QOS = ∠QOR + ∠ROS

∠QOS = 90° + ∠ROS

∴ ∠ROS = ∠QOS – 90° (ii)

By adding equation (i) and (ii)

∴ ∠ROS = (frac{1}{2}) (∠QOS – ∠POS)

Question 6.

It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYO and reflex ∠QYP.

Answer:

Data: ∠XYZ = 64° and XY are produced up to P. ∠ZYP is bisected.

To Prove: ∠XYQ = ? Reflex ∠QYP = ?

Proof: YQ bisects ∠ZYP

∴ Let ∠PYQ = ∠ZYQ = x°.

∠XYZ + ∠ZYQ + ∠QYP = 180° (∵ ∠XYP is straight angle)

64 + x + x = 180

∴ 64 + 2x = 180

∴ 2x = 180 – 64

2x = 116

∴ (x=frac{116}{2}=58^{circ})

∴ ∠PYQ = ∠ZYQ = 58°

(i) ∴ ∠XYQ = ∠XYZ + ∠ZYQ

= 64 + 58

∠XYQ = 122°

(ii) Reflex ∠QYP = ∠PYX + ∠XYZ + ∠ZYQ

= 180 ° + 64° + 58°

= 180 + 122

∴ Reflex ∠QYP = 302°.

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