Concept Of Vertically Opposite Angles

Vertically Opposite Angles:

Let us understand the meaning of these words.

Vertically opposite means anything which is lying on the other side. So vertically opposite angles refer to those set of angles which are opposite to each other, in other words lying face to face to each other.

Now for this question it is given that PQ and RS are two straight lines.

So, these two lines intersect at a point O in the given manner to form a cross:

Now when these two lines intersect, two pairs of vertically opposite angles are formed.

Now from the above figure,

∠POS is lying opposite to ∠ROQ

So, our first pair of vertically opposite angles are ∠POS and ∠ROQ.

Now when we again observe the figure, we notice that ∠POR and ∠SOQ are also lying opposite to each other.

So, our second pair of vertically opposite angles is ∠POR and ∠SOQ.

Additional Information:

Vertically opposite angles are always equal.

Like in above figure, ∠POS = ∠ROQ

∠POR = ∠SOQ

Now when we observe all the angles, it forms a circle. Hence the sum of all vertically opposite angles is always 360°.

In the above figure,

∠POS + ∠ROQ + ∠POR + ∠SOQ = 360°

In the above given figure, let us understand the pairs of vertically opposite angle.

∠1 and ∠3 are lying opposite to each other [in common language lying face to face to each other].

So, first pair of vertically opposite angles is ∠1 and ∠3.

∠4 and ∠2 are also lying opposite to each other.

Hence ∠4 and ∠2 forms our second pair of vertically opposite angles.

Vertically opposite angles are always equal.

Hence ∠1 must be equal to ∠3 and ∠4 must be equal to ∠2.

Also, we know that sum all vertically opposite angles is 360°.

∠1 + ∠2 + ∠3 + ∠4 = 360°.

Now in the question it is given that,

∠1 = 35°

So ∠3 must be equal to 35° [as ∠3 = ∠1……… vertically opposite angles]

Let ∠2 = ∠4 = x [as ∠4 = ∠2……… vertically opposite angles]

So, 35° + 35° + x + x = 360°

70° + 2x = 360°

2x = 360 – 70

2x = 290

x = 290 / 2

x = 145°

So ∠2 = ∠4 = 145°.

In the above given figure, let us understand the pairs of vertically opposite angle.

∠ROQ is lying opposite to ∠POS.

So, our first pair of vertically opposite angles is ∠ROQ and ∠POS.

∠ROP is lying opposite to ∠QOS.

So, our second pair of vertically opposite angles is ∠ROP and ∠QOS.

Vertically opposite angles are equal.

So ∠ROQ = ∠POS

and ∠ROP = ∠QOS.

Sum of all vertically opposite angles is 360°.

∠ROQ + ∠POS + ∠ROP + ∠QOS = 360°.

Now in the above question,

∠POS = ∠POT + ∠TOS

∠ROQ = 60°.

∠TOS = 20°

So ∠POS = ∠ROQ = 60°.

∠POT = ∠POS - ∠TOS

= 60 – 20

= 40°.

Let ∠ROP = ∠QOS = x [as ∠QOS = ∠ROP……… vertically opposite angles]

So, 60° + 60° + x + x = 360°

120° + 2x = 360°

2x = 360 – 120

2x = 240

x = 240 / 2

x = 120°

So ∠QOS = ∠ROP = 120°.

Let us consider the two lines to be drawn in following way:

Let ∠POX = ∠1 and ∠YOQ = ∠3 and ∠POY = ∠4 and ∠QOX = ∠2.

Now let us measure the angles one by one with help of protractor.

So ∠1 = 60°

∠2 = 120°

∠3 = 60°

∠4 = 120°

After measuring these angles, we can notice that the opposite angles are equal.

That is ∠1 = ∠3 = 60°

∠2 =∠4 = 120°

One more thing which can be noticed is that sum of all the angles is 360°.

That is ∠1 + ∠2 + ∠3 + ∠4 = 360°.

i. Complementary angles is defined as those pairs of angles which sum up to 90°.

In the above figure, OM is the angle bisector.

Ray OD is perpendicular to line AB.

So ∠AOM + ∠MOD = 90°.

Hence ∠AOM and ∠MOD are two complementary angles.

ii. Supplementary angles can be defined as those pairs of angles which sum up to 180°.

In the above figure,

∠AOD = 90° = ∠BOD ………. (as OD is perpendicular to line AB)

So ∠AOD + ∠BOD = 180°.

Similarly ray CO is perpendicular to line AB.

So ∠AOC + ∠BOC = 180°

So, the two pairs of supplementary angles are ∠AOD & ∠BOD and ∠AOC & ∠BOC

iii. In the above figure,

AB and CD are two straight lines intersecting at point O.

∠AOD is lying opposite to ∠BOC.

Similarly, ∠BOD is lying opposite to ∠AOC.

Hence ∠AOD and ∠BOC form the first pair of vertically opposite angles.

∠AOC and ∠BOD forms the second pair of vertically opposite angles.

Let us consider two straight lines PQ and RS intersecting at a point O.

Let ∠POR = ∠1

∠ROQ = ∠2

∠SOQ = ∠3

∠SOP = ∠4

Let us first find the pairs of vertically opposite angles from above diagram.

∠1 and ∠3 are lying opposite to each other [in common language lying face to face to each other].

So, first pair of vertically opposite angles is ∠1 and ∠3.

∠4 and ∠2 are also lying opposite to each other.

Hence ∠4 and angle 2 forms our second pair of vertically opposite angles.

To Prove: Vertically opposite angles are equal

Proof:

Now when we observe the above diagram, ray RO is perpendicular to line PQ.

So ∠1 + ∠2 = 180°……… (i) [As they form linear pair that is sum of angles in a straight line is 180°]

Similarly ray QO is perpendicular to line SR.

So ∠2 + ∠3 = 180°……… (ii) [As they form linear pair that is sum of angles in a straight line is 180°]

Now for equation (i) and (ii), the right-hand side is same. So, both the terms of left-hand side have to be equal.

That is ∠1 + ∠2 = ∠2 + ∠3

∠2 is eliminated and the final equation becomes,

∠1 = ∠3.

Hence it is proved that ∠1 and ∠3 are equal and at the same time they are vertically opposite angles.

Now ray PO is perpendicular to line SR.

So ∠4 + ∠1 = 180°……… (iii) [As they form linear pair that is sum of angles in a straight line is 180°]

Now for equation (i) and (iii), the right-hand side is same. So, both the terms of left-hand side has to be equal.

That is ∠1 + ∠2 = ∠4 + ∠1

∠1 is eliminated and the final equation becomes,

∠2 = ∠4.

Hence it is proved that ∠2 and ∠4 are equal and at the same time they are vertically opposite angles.

In the above given figure, let us understand the pairs of vertically opposite angle.

∠AOD and ∠COB are lying opposite to each other [in common language lying face to face to each other].

So, first pair of vertically opposite angles is ∠AOD and ∠COB.

∠AOC and ∠DOB are also lying opposite to each other.

Hence ∠AOC and ∠DOB forms our second pair of vertically opposite angles.

Vertically opposite angles are always equal.

Hence ∠AOD must be equal to ∠COB and ∠AOC must be equal to ∠DOB.

Also, we know that sum all vertically opposite angles is 360°.

∠AOD + ∠COB + ∠AOC + ∠DOB = 360°.

Now in the question it is given that,

∠AOD = 120°

So ∠COB must be equal to 120° [as ∠COB = ∠AOD……… vertically opposite angles]

Let ∠DOB = ∠AOC = x [as ∠AOC = ∠DOB……… vertically opposite angles]

So, 120° + 120° + x + x = 360°

240° + 2x = 360°

2x = 360 – 240

2x = 120

x = 120 / 2

x = 60°

So ∠AOC = ∠DOB = 60°.

Let us first find out the pairs of vertically opposite angles from the above given figure.

Now from the above figure,

∠POS is lying opposite to ∠ROQ

So, our first pair of vertically opposite angles are ∠POS and ∠ROQ.

Now when we again observe the figure, we notice that ∠POR and ∠SOQ are also lying opposite to each other.

So, our second pair of vertically opposite angles is ∠POR and ∠SOQ.

Now we know two basic properties of vertically opposite angles which are as follows:

1. Vertically opposite angles are always equal.

So, we can say that ∠POS = ∠ROQ and ∠POR = ∠SOQ

2. Sum of all vertically opposite angles is 360°.

∠POS + ∠ROQ + ∠POR + ∠SOQ = 360° ……... (i)

Now in the question it given that sum of angles ∠POR and ∠SOQ is 110°.

Let ∠POR = ∠SOQ = x [as these both angles are equal]

So, x + x = 110°

2x = 110°

x = 110 / 2

x = 55°

So ∠POR = ∠SOQ = 55°.

Let ∠POS = ∠ROQ = x [as these both angles are equal]

Let us substitute the values in equation (i)

55° + 55° + x + x = 360°

110° + 2x = 360°

2x = 360 – 110

2x = 250

x = 250 / 2

x = 125°

∠POS = ∠ROQ = 125°.

Now let us first construct a diagram from the explanation given in the question.

It is mentioned that OP, OQ, OR and OS are concurrent.

Concurrent means multiple line intersecting at a single common point.

If we carefully look at these rays, each ray has common point O. So, the lines intersect at a common point O.

Now it is given that OP and OR is on same straight line. At the same time P and R are situated on opposite sides of point O.

So, our first straight line is PR which intersecting with line SQ at a common point O.

In the above given figure, let us understand the pairs of vertically opposite angle.

∠POQ and ∠SOR are lying opposite to each other [in common language lying face to face to each other].

So, first pair of vertically opposite angles is ∠POQ and ∠SOR.

∠POS and ∠QOR are also lying opposite to each other.

Hence ∠POS and ∠QOR forms our second pair of vertically opposite angles.

Let ∠POQ = ∠1

∠QOR = ∠2

∠SOR = ∠3

∠POS = ∠4.

Vertically opposite angles are always equal.

Hence ∠1 must be equal to ∠3 and ∠4 has to be equal to ∠2.

Also, we know that sum all vertically opposite angles is 360°.

∠1 + ∠2 + ∠3 + ∠4 = 360°.

Now in the question it is given that,

∠1 = 50°

So ∠3 must be equal to 50° [as ∠1 = ∠3……… vertically opposite angles]

Let ∠4 = ∠2 = x [as ∠2 = ∠4……… vertically opposite angles]

So, 50° + 50° + x + x = 360°

100° + 2x = 360°

2x = 360 – 100

2x = 260

x = 260 / 2

x = 130°

So ∠2 = ∠4 = 130°.

So ∠POQ = ∠SOR = 50°.

∠POS = ∠QOR = 130°.

Let us consider four rays, OP, OQ, OR and OS meet at a common point.

Now in these rays the common point is O. Hence all the rays meet at a common point O.

Now in the above figure rays meet at a common point O.

It is given that opposite angles are always equal.

In the above given figure, let us understand the pairs of vertically opposite angle.

∠POQ and ∠SOR are lying opposite to each other [in common language lying face to face to each other].

So, first pair of vertically opposite angles is ∠POQ and ∠SOR.

∠POS and ∠QOR are also lying opposite to each other.

Hence ∠POS and ∠QOR forms our second pair of vertically opposite angles.

To Prove: Four rays form two straight lines

Proof:

Let ∠POQ = ∠1

∠QOR = ∠2

∠SOR = ∠3

∠POS = ∠4.

Now when we closely look at the figure,

∠1 = ∠3 and ∠2 = ∠4.

If we add all these angles it will be sum up to 360° as the angles form a complete circle.

Let ∠1 = ∠3 = x and ∠2 = ∠4 = y

So ∠1 + ∠2 + ∠3 + ∠4 = 360°

2x + 2y = 360°

2(x + y) = 360°

x + y = 180°

The above equation suggests that sum of any two side angles is 180° that is ∠1 + ∠2 = 180°

Or ∠2 + ∠3 = 180°

Or ∠3 + ∠4 = 180°

Or ∠1 + ∠4 = 180°

The sum of two side angles is 180° which means that they form a linear pair.

Linear Pair case is possible only when two straight lines intersect at a common point.

Hence it is proved that four rays intersect to form two straight lines.

Let us first draw a figure to support our proof.

Proof:

In the above figure PQ is a straight line. O is the point of intersection of line with ray RO, TO and SO.

Let ∠POS = x and let ∠QOS = y

Here ∠POS is an internal angle and ∠QOS is an external angle.

Since PQ is a straight line,

x + y = 180°

x = 180 – y ……. (i)

Let OR be the internal angle bisector.

So ∠POR = ∠SOR = 1/2 ∠POS [as angle bisector divides the given angle in two halves]

In the similar manner OT is the external angle bisector.

So, we can write ∠SOT = ∠TOQ = 1/2 ∠SOQ

Now when we combine, we get a common angle bisector ∠ROT.

∠ROT = ∠SOR + ∠SOT

∠ROT = 1/2 ∠POS + 1/2 ∠QOS

= 1/2 × (∠POS + ∠QOS)

= 1/2 × (x + y)

= 1/2 × (180 – y + y) ………. (From equation (i))

= 1/2 × 180

∠ROT = 90°

Since the angle is 90°, it is proved that internal and external bisectors of an angle are perpendicular to each other.

Let us consider two straight lines PR and SQ intersecting at a common point O.

The question is just a twisted version of very simple question.

Proof:

We know that a right angle means that the angle is equal to 90°.

So, four right angles mean that the angle is equal to 4 × 90° which is equal to 360°.

Let ∠POQ = ∠1

∠QOR = ∠2

∠SOR = ∠3

∠POS = ∠4.

Now when we observe the above diagram, ray QO is perpendicular to line PR.

So ∠1 + ∠2 = 180°……… (i) [As they form linear pair that is sum of angles in a straight line is 180°]

Similarly ray RO is perpendicular to line SQ.

So ∠2 + ∠3 = 180°……… (ii) [As they form linear pair that is sum of angles in a straight line is 180°]

Now let us add equations (i) and (ii).

So, we get ∠1 + ∠2 + ∠3 + ∠4 = 180° + 180°

∠1 + ∠2 + ∠3 + ∠4 = 360°.

Hence all the angles sum up to four right angles that is 360°.

Hence it is proved that the sum of measurement of the four angles is four right angles.

Let us first construct diagram for the above proof:

QR is extended on both the sides.

In the figure, ∠PQS and ∠PRT are the two exterior angles.

It is given that ∠PQR = ∠PRQ which means that the given triangle is an isosceles triangle.

Hence length PQ = PR.

Now ∠PQS + ∠PQR = 180° ……(i) [as they form linear pair that angles in straight line always add up to 180°]

Similarly, ∠PRQ + ∠PRT = 180° ……. (ii).

For the equations (i) and (ii), the right-hand side is equal which means that the left-hand side should also be equal.

∠PQS + ∠PQR = ∠PRQ + ∠PRT

∠PQS and ∠PRQ gets eliminated as they both are equal angles.

So, we can state that ∠PQS = ∠PRT.

Hence it is proved that the measurement of external angles is equal.

Let us first draw a diagram for the above question.

In the above figure,

PQ and SR are two straight lines.

O is the common point.

AO is the angle bisector of ∠POS.

NO is the angle bisector of ∠SOQ

BO is the angle bisector of ∠QOR

MO is the angle bisector of ∠POR.

∠POR = ∠SOQ [vertically opposite angles are always equal]

Similarly, ∠POS = ∠QOR [vertically opposite angles are always equal]

Let ∠POR = ∠SOQ = x and Let ∠POS = ∠QOR = y.

OR is perpendicular to line PQ.

So ∠POR + ∠QOR = 180° [as they form linear pair that is angles in a straight line always sum up to 180°]

∠MOR = ∠MOP = 1/2 ∠POR [as MO is the angle bisector]

Similarly, ∠BOR = ∠QOB = 1/2 ∠QOR [as BO is the angle bisector]

∠MOB = ∠MOR + ∠BOR

= 1/2 ∠POR + 1/2 ∠QOR

= 1/2 × (∠POR + ∠QOR)

= 1/2 × 180

∠MOB = 90°

So ∠SOQ + ∠QOR = 180° [as they form linear pair that is angles in a straight line always sum up to 180°]

∠NOS = ∠NOQ = 1/2 ∠SOQ [as NO is the angle bisector]

Similarly, ∠QOB = ∠BOR = 1/2 ∠QOR [as BO is the angle bisector]

∠NOB = ∠NOQ + ∠QOB

= 1/2 ∠SOQ + 1/2 ∠QOR

= 1/2 × (∠SOQ + ∠QOR)

= 1/2 × 180

∠NOB = 90°

Since the angle is 90°, it is proved that the bisectors of these angles are two perpendicular straight lines.