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9th class > Lines and Angles > Pairs of Angles

Pairs of Angles

Concerning angles, we know about complementary angles, supplementary angles, adjacent angles, linear pair of angles etc. Now, let's move on to finding out about angles formed when a ray is placed on a line.

PQ Line with QS ray

What are the angles formed at the point Q? They are ∠ PQS, ∠ SQR and ∠ PQR.

Can we write ∠ PQS + ∠ SQR = ∠ PQR? (1)

What is the measure of ∠ PQR? It is ° (2)

From (1) and (2), we can now say that-

∠ PQS + ∠ SQR = 180°

From here, we can state the following Axiom:

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

Since, here the sum of measures of the two adjacent angles is equal to 180°, they are together called a linear pair of angles.

Now, is the converse of this statement true? i.e.

If the sum of two adjacent angles is 180°, then a ray stands on a line aka the non-common arms form a line?

Upon examining multiple pairs of angles with different mesures- as it turns out, it is in fact true. Thus,

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

The two axioms that we have just studied together are known as the Linear Pair Axiom.

Moving on to the next situation where two lines intersect at a common point. We know that when two lines intersect, the vertically opposite angles are equal. So, let's prove it now.

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

AB and CD intersecting at O

Proof : Let lines AB and CD intersect at O as shown above. This leads to the formation of two pairs of vertically opposite angles

(i) ∠ AOC and ∠ BOD

(ii) ∠ AOD and ∠ BOC

We need to prove that: ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC

Instructions

Check the above figure

  • We can say that OA stands on line CD
  • Using linear pair axiom we get
  • Similarly, ray stands on line AB
  • Again using linear pair axiom we get
  • From (1) and (2) together
  • Thus, we get: ∠ AOC = ∠
  • Similarly, we can conclude: ∠AOD = ∠

Example 1 : In the below figure, lines PQ and RS intersect each other at point O. If ∠ POR : ∠ ROQ = 5 : 7, find all the angles.

Figure for example 1

Solution : ∠ POR + ∠ROQ = ° (Linear pair of angles)

But ∠ POR : ∠ ROQ = 5 : 7 (Given)

Therefore, ∠ POR = × = °

Similarly, ∠ ROQ = × ° = °

Now, ∠ POS = = ° ( angles)

and ∠ SOQ = = ° (Vertically opposite angles)

Example 2: In the below figure, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠ POS and ∠ SOQ, respectively. If ∠ POS = x, the ∠ ROT is equal to :

Instructions

Ray OS stands on line POQ

  • By linear pair axiom: ∠ POS + ∠ SOQ = °
  • As ∠ POS = x we get: ∠ SOQ = 180° – x
  • Given that: OR and OT bisect the angles ∠ POS and ∠ SOQ.
  • Thus, ∠ ROS = ° while ∠ SOT = °
  • We see that: ∠ ROT = ∠ ROS + ∠
  • Thus, ∠ ROT = °
  • Substituting values
  • We have found the answer

Example 3 : In the given figure, OP, OQ, OR and OS are four rays. Prove that ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°.

Figure for example 3

Solution : In the figure, you 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 (as shown below).

Modified figure for example 3

Now, ray OP stands on line TOQ.

Therefore, ∠TOP + = ° (1) (Linear pair axiom)

Similarly, ray OS stands on line TOQ.

Therefore, ∠TOS + = ° (2)

But ∠ SOQ = ∠ SOR +

So, (2) becomes ∠ TOS + ∠ SOR + ∠ QOR = ° (3)

Now, adding (1) and (3), you get:

∠ TOP + ∠ POQ + ∠ TOS + ∠ SOR + ∠ QOR = ° (4)

But ∠ TOP + ∠ TOS = ∠

Therefore, (4) becomes ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = °