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NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Exercise 6.1

Q1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and AnglesAns: From the diagram, we have
(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.
So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°
Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get
∠COE = 110° and ∠BOE = 30°
So, reflex ∠COE = 360°  – 110° = 250°


Q2. In the following figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Ans:We know that the sum of linear pair are always equal to 180°
So,
POY +a +b = 180°
Putting the value of POY = 90° (as given in the question) we get,
a+b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x
∴ 2x+3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2×18° = 36°
Similarly, b can be calculated and the value will be
b = 3×18° = 54°
From the diagram, b+c also forms a straight angle so,
b+c = 180°
c+54° = 180°
∴ c = 126°


Q3. In the following figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Ans: Since ST is a straight line so,
∠PQS+∠PQR = 180° (linear pair) and
∠PRT+∠PRQ = 180° (linear pair)
Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°
Since ∠PQR =∠PRQ (as given in the question)
∠PQS = ∠PRT. (Hence proved).


Q4. In the following figure, if x + y = w + z, then prove that AOB is a line.

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Ans: For proving AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x+y = 180°
We know that the angles around a point are 360° so,
x + y + w + z = 360°
In the question, it is given that,
x+y = w+z
So, (x+y)+(x+y) = 360°
2(x+y) = 360°
∴ (x+y) = 180° (Hence proved).


Q5. In the adjoining figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½ (∠QOS – ∠POS).

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Ans: In the question, it is given that (OR ⊥ PQ) and POQ = 180°
So, POS+ROS+ROQ = 180°
Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°)
∴ POS + ROS = 90°
Now, QOS = ROQ+ROS
It is given that ROQ = 90°,
∴ QOS = 90° +ROS
Or, QOS – ROS = 90°
As POS + ROS = 90° and QOS – ROS = 90°, we get
POS + ROS = QOS – ROS
2 ROS + POS = QOS
Or, ROS = ½ (QOS – POS) (Hence proved).


Q6. 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 ∠XYQ and reflex ∠QYP.
Ans:

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and AnglesHere, XP is a straight lineSo, XYZ +ZYP = 180°
Putting the value of XYZ = 64° we get,
64° +ZYP = 180°
∴ ZYP = 116°
From the diagram, we also know that ZYP = ZYQ + QYP
Now, as YQ bisects ZYP,
ZYQ = QYP
Or, ZYP = 2ZYQ
∴ ZYQ = QYP = 58°
Again, XYQ = XYZ + ZYQ
By putting the value of XYZ = 64° and ZYQ = 58° we get.
XYQ = 64°+58°
Or, XYQ = 122°
Now, reflex QYP = 180°+XYQ
We computed that the value of XYQ = 122°.
So,
QYP = 180°+122°
∴ QYP = 302°


Exercise 6.2

Q1. In the following Figure, if AB CD, CD EF and y : z = 3 : 7, find x.

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Ans: It is known that ABCD and CDEF
As the angles on the same side of a transversal line sums up to 180°,
x + y = 180° —–(i)
Also,
O = z (Since they are corresponding angles)
and, y +O = 180° (Since they are a linear pair)
So, y+z = 180°
Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7)
∴ 3w+7w = 180°
Or, 10 w = 180°
So, w = 18°
Now, y = 3×18° = 54°
and, z = 7×18° = 126°
Now, angle x can be calculated from equation (i)
x+y = 180°
Or, x+54° = 180°
∴ x = 126°


Q2. In the following figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and AnglesAns: Since AB || CD, GE is a transversal.
It is given that ∠GED = 126°
So, ∠GED = ∠AGE = 126° (As they are alternate interior angles)
Also,
∠GED = ∠GEF +∠FED
As EF⊥ CD, ∠FED = 90°
∴ ∠GED = ∠GEF+90°
Or, ∠GEF = 126° – 90° = 36°
Again, ∠FGE +∠GED = 180° (Transversal)
Putting the value of ∠GED = 126°, we get
∠FGE = 54°
So,
∠AGE = 126°
∠GEF = 36° and
∠FGE = 54°


Q3. In the following figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Ans: First, construct a line XY parallel to PQ.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles
We know that the angles on the same side of transversal is equal to 180°.
So, ∠PQR+∠QRX = 180°
Or, ∠QRX = 180°-110°
∴ ∠QRX = 70°
Similarly,
∠RST +∠SRY = 180°
Or, ∠SRY = 180°- 130°
∴ ∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX+∠QRS+∠SRY = 180°
Putting their respective values, we get
∠QRS = 180° – 70° – 50°
Hence, ∠QRS = 60°


Q4. In the following figure, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and AnglesAns: From the diagram,
∠APQ = ∠PQR (Alternate interior angles)
Now, putting the value of ∠APQ = 50° and ∠PQR = x, we get
x = 50°
Also,
∠APR = ∠PRD (Alternate interior angles)
Or, ∠APR = 127° (As it is given that ∠PRD = 127°)
We know that
∠APR = ∠APQ+∠QPR
Now, putting values of ∠QPR = y and ∠APR = 127°, we get
127° = 50°+ y
Or, y = 77°
Thus, the values of x and y are calculated as:
x = 50° and y = 77°


Q5. In the following figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles
Ans: First, draw two lines, BE and CF, such that BE ⊥ PQ and CF ⊥ RS.
Now, since PQ || RS,
So, BE || CF
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

We know that,
Angle of incidence = Angle of reflection (By the law of reflection)
So,
∠1 = ∠2 and
∠3 = ∠4
We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C
So, ∠2 = ∠3 (As they are alternate interior angles)
Now, ∠1 +∠2 = ∠3 +∠4
Or, ∠ABC = ∠DCB
So, AB || CD (alternate interior angles are equal)

The document NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

1. What are lines and angles?
Ans. Lines and angles are fundamental concepts in geometry. A line is a straight path that extends infinitely in both directions. An angle is formed when two lines or line segments meet at a common point called the vertex.
2. How do you classify angles based on their measures?
Ans. Angles can be classified into different types based on their measures. They can be classified as acute angles (less than 90 degrees), right angles (exactly 90 degrees), obtuse angles (more than 90 degrees but less than 180 degrees), and straight angles (exactly 180 degrees).
3. What is the relationship between parallel lines and transversals?
Ans. When a transversal intersects two parallel lines, several pairs of angles are formed. These pairs of angles can be classified into corresponding angles, alternate interior angles, alternate exterior angles, and consecutive interior angles.
4. How can we determine if two lines are perpendicular?
Ans. Two lines are perpendicular to each other if they intersect at a right angle (90 degrees). In other words, the angles formed by the intersection of two lines are right angles.
5. What is the sum of the angles in a triangle?
Ans. The sum of the angles in a triangle is always 180 degrees. This property is known as the angle sum property of triangles.
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