NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.

In Section 6.2, we learnt the definitions of some of the pairs of angles such as complementary angles, supplementary angles, adjacent angles, linear pair of angles, etc. Can we think of some relations between these angles?

Now, let us find out the relation between the angles formed when a ray stands on a line. Draw a figure in which a ray stands on a line as shown in Figure 6.6.

Figure 6.6 : Linear pair of angles

 

Let us name the line as AB and the ray as OC. What are the angles formed at the point O?

They are

\[∠AOC,\ ∠BO, ∠AOB.\]

Can we write

\[∠AOC\ +\ ∠BOC\ =\ ∠AOB?\ ...\ ...\ ...\ (1)\]

Yes! (Why? Refer to adjacent angles in Section 6.2)

What is the measure of \(∠AOB?\) It is 180°. (Why?) ... ... ... (2)

From (1) and (2), can we say that \(∠AOC\ +\ ∠BOC\ =\ 180°?\) Yes! (Why?)

From the above discussion, we can state the following Axiom:

Axiom 6.1

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Recall that when the sum of two adjacent angles is 180°, then they are called a linear pair of angles.

In Axiom 6.1, it is given that ‘a ray stands on a line’. From this ‘given’, we have concluded that ‘the sum of two adjacent angles so formed is 180°’. Can we write Axiom 6.1 the other way round? That is, take the ‘conclusion’ of Axiom 6.1 as ‘given’ and the ‘given’ as the ‘conclusion’. So, it becomes:

statement (A)
If the sum of two adjacent angles is 180°, then a ray stands on a line (that is, the non-common arms form a straight line).

Now we see that the Axiom 6.1 and statement (A) are in a sense the reverse of each other. We call each as converse of the other. We do not know whether the statement (A) is true or not. Let us check.

Draw adjacent angles of different measures as shown in Figure 6.7. Keep the ruler along one of the non-common arms in each case. Does the other non-common arm also lie along the ruler?

Figure 6.7(i) : Adjacent angles with different measures

 

Figure 6.7(ii) : Adjacent angles with different measures

 

Figure 6.7(iii) : Adjacent angles with different measures

 

Figure 6.7(iv) : Adjacent angles with different measures

 

We will find that only in Figure 6.7 (iii), both the non-common arms lie along the ruler, that is, points A, O and B lie on the same line and ray O C stands on it. Also see that

\[∠AOC\ +\ ∠COB\ =\ 125°\ +\ 55°\ =\ 180°.\]

From this, we may conclude that statement (A) is true. So, we can state in the form of an axiom as follows:

Axiom 6.2

If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.

For obvious reasons, the two axioms above together is called the Linear Pair Axiom.

Let us now examine the case when two lines intersect each other.

Let us recall, from our earlier classes, that when two lines intersect, the vertically opposite angles are equal. Let us prove this result now. See Appendix 1 for the ingredients of a proof and keep those in mind while studying the proof given below.

Theorem 6.1

If two lines intersect each other, then the vertically opposite angles are equal.

Proof :

In the statement above, it is given that ‘two lines intersect each other’. So, let AB and CD be two lines intersecting at Point O as shown in Figure 6.8.

Figure 6.8 : Vertically opposite angles

 

They lead to two pairs of vertically opposite angles, namely,

\[(i)\ ∠AOC,\ ∠BOD\]

\[(ii)\ ∠AOD,\ ∠BOC.\]

We need to prove that

\[∠AOC\ =\ ∠BOD;\ ∠AOD\ =\ ∠BOC.\]

Now, ray OA stands on line CD.

\[∴\ ∠AOC\ +\ ∠AOD\ =\ 180°\]
(Linear pair axiom) ... ... (1)

Can we write

\[∠AOD\ +\ ∠BOD\ =\ 180°?\]
Yes! (Why?) ... ... (2)

From (1) and (2), we can write

\[∠AOC\ +\ ∠AOD\ =\ ∠AOD\ +\ ∠BOD\]

This implies that

\[∠AOC\ =\ ∠BOD\
(Refer Section 5.2, Axiom 3)

Similarly, it can be proved that

\[∠AOD\ =\ ∠BOC\]

Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1.

Example 1

In Figure 6.9, line PQ and Line RS intersect each other at point O. If

\[∠POR\ ∶\ ∠ROQ\ =\ 5\ ∶\ 7,\]

find all the angles.

Figure 6.9

 

Solution :

\[∠POR\ +\ ∠ROQ\ =\ 180°\]
(Linear pair of angles)

\[∵\ ∠POR\ ∶\ ∠ROQ\ =\ 5\ ∶\ 7\]

\[∴\ ∠POR\ =\ \frac{5}{12}\ ×\ 180°\ =\ 75°\]

Similarly,

\[∠ROQ\ =\ \frac{7}{12}\ ×\ 180°\ =\ 105°\]

Now,

\[∠POS\ =\ ∠ROQ\ =\ 105°\]
(Vertically opposite angles)

and

\[∠SOQ\ =\ ∠POR\ =\ 75°\]
(Vertically opposite angles)

Example 2

In Figure 6.10, Ray OS stands on Line PQ. Ray OR and Ray OT are ∠bisectors of ∠POS and ∠SOQ, respectively. If ∠POS = x, find ∠ROT.

Figure 6.10

 

Solution :

Ray OS stands on the line PQ, where Point O is on the Line AB.

\[∴\ ∠POS\ +\ ∠SOQ\ =\ 180°\]

\[∵\ ∠POS = x\]

\[∴,\ x\ +\ ∠SOQ\ =\ 180°\]

\[∴,\ ∠SOQ\ =\ 180°\ −\ x\]

Now, ray OR bisects ∠POS, therefore,

\[∴,\ ∠ROS\ =\ ½\ ×\ ∠POS\]

\[=\ ½\ ×\ x\ =\ x/2\]

Similarly,

\[∠SOT\ =\ ½\ ×\ ∠SOQ\]

\[=\ ½\ ×\ (180°\ −\ x)\]

= 90° − x/2

Now,

\[∠ROT\ =\ ∠ROS\ +\ ∠SOT\]

\[=\ x/2\ +\ 90°\ −\ x/2\]

= 90°

Example 3

In Figure 6.11, oP, oQ, oR and OS are four rays. Prove that

\[∠POQ\ +\ ∠QOR\ +\ ∠SOR\ +\ ∠POS\ =\ 360°.\]

Figure 6.11

 

Solution :

In Figure 6.11, we need to produce any of the rays oP, oQ, oR or OS backwards to a point. Let us produce ray oQ backwards to a point T so that TOQ is a line (see Figure 6.12).

Figure 6.12

 

Now, ray oP stands on line TOQ.

\[∴,\ ∠TOP\ +\ ∠POQ\ =\ 180°\ ...\ ...\ (1)\]
(Linear pair axiom)

Similarly, ray OS stands on line TQ.

\[∴,\ ∠TOS\ +\ ∠SOQ\ =\ 180°\ ...\ ...\ (2)\]

\[∵\ ∠SOQ\ =\ ∠SOR\ +\ ∠QOR\]

So, (2) becomes

\[∠TOS\ +\ ∠SOR\ +\ ∠QOR\ =\ 180°\ ...\ ...\ (3)\]

Now, adding (1) and (3), we get

\[∠TOP\ +\ ∠POQ\ +\ ∠TOS\ +\ ∠SOR\ +\ ∠QOR\ =\ 360°\ ...\ ...\ (4)\]

\[∵\ ∠TOP\ +\ ∠TOS\ =\ ∠POS\]

Therefore, (4) becomes

\[∠POQ\ +\ ∠QOR\ +\ ∠SOR\ +\ ∠POS\ =\ 360°\]

EXERCISE 6.1

Q1. In Figure 6.13, Line AB and Line CD intersect at Point O. If
\[∠AOC\ +\ ∠BOE\ =\ 70°\]
and
\[∠BOD\ =\ 40°,\]

find ∠BOE and reflex angle COE.

Figure 6.13

 

A1. 30°, 250°

Q2. In Figure 6.14, Line XY and Line MN intersect at Point O. If
\[∠POY\ =\ 90°\ \ a\ ∶\ b\ =\ 2\ ∶\ 3,\]
find c.

Figure 6.14

 

A2. 126°

Q3. In Figure 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Figure 6.15

 

Q4. In Figure 6.16, if x + y = w + z, then prove that AOB is a line.

Figure 6.16

 

A4. Sum of all the angles at a point = 360°

Q5. In Figure 6.17, POQ is a line. Ray oR is perpendicular to line PQ. OS is another ray lying between rays oP and oR. Prove that
\[∠ROS\ =\ ½\ ×(∠QOS − ∠POS).

Figure 6.17

 

A5.
\[∠QOS\ =\ ∠SOR\ +\ ∠ROQ;\ ∠POS\ =\ ∠POR\ −\ ∠SOR.\]

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.

A6. 122°, 302°