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°.
All Chapter KSEEB Solutions For Class 9 Maths
—————————————————————————–
All Subject KSEEB Solutions For Class 9
*************************************************
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.
Best of Luck!!