WBBSE Solutions For Class 8 Maths Chapter 7 Concept Of Vertically Opposite Angles

Concept Of Vertically Opposite Angles

It is seen that two pieces of ribbons on the card of Motiur are in one way while the two pieces of ribbons on the card of Annesha are in another way. However, the ribbons in the two cards intersect each other and form several angles. Let’s find the relation between them.

Measuring by protractor we see that in the first card ∠1 = 90 degree ∠2 = 90 degree ∠3 = 90 degree ∠4 = 90 degree

∠1= ∠2= ∠3= ∠4= 90 degree

“WBBSE Class 8 Maths Chapter 7 solutions, Concept of Vertically Opposite Angles”

Measuring by protractor we see that in the ∠1 = 120 degree ∠2 = 60 degree ∠3 = 120 degree ∠4 = 60 degree

Also, we see by measuring by protractor ∠1 =∠3 =120 degree ∠ 2 = ∠4 = 60 degree

Read and Learn More WBBSE Solutions For Class 8 Maths

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

(Given): AB and CD intersect at O. As a result two pairs of vertically opposite angles ∠AOD, ∠BOG, and ∠AOC, ∠BOD are formed.

∠AOD = ∠BOC and ∠AOC = ∠BOD

Proof :

∠AOD + ∠AOC = 180° [As CD stands on OA, therefore the sum of the measurement of angles is 180°]

∠AOC + ∠BOC = 180° [As AB stands on OC, therefore the sum of measurement of adjacent angles is 180°

∠AOD + ∠AOC = ∠AOC + ∠BOC Hence, ∠AOD = ∠BOC ( Subtracting ∠AOC from both sides). Again, we can write-

BOC + BOD = 180° [As AB stands on CD, therefore the sum of measurement of adjacent angles is 2 right angles.]

Concept Of Vertically Opposite Angles Exercise

Question 1.  Two straight lines PQ and RS intersect at O and some vertically opposite angles are formed, let’s draw and write the names.

Given

Two straight lines PQ and RS intersect at O and some vertically opposite angles are formed

1. Let’s try to find the measurement of the angles from the figure below.

Solution:

1. ∠1=35°
2. ∠2=145°
3. ∠3=35°
4. ∠4=145°

2. 

1. ∠TOS=20°
2. ∠ROQ=60°
3. ∠POT=40°
4. ∠ROP=120°
5. ∠QOS=120°

Question 3. Tirtha drew two straight lines PQ and XY intersecting at O. Let’s measure the vertically opposite angles by a protractor.

Solution:

Given

Tirtha drew two straight lines PQ and XY intersecting at O.

∠POX = opposite ∠YOQ = 50c

or, ∠POY = opposite ∠QOX= 130°

Question 4. Let’s try to find the answer to the questions given below, studying the figure beside.

Solution:

1. ∠AOM and ∠MOD are complementary angles
2. ∠AOC and ∠BOC are complementary angles
3. ∠AOC and ∠BOD are complementary angles

Question 5. If two straight lines intersect each other then the measurement of vertically opposite angles are equal; prove it logically.

Solution:

Given

If two straight lines intersect each other then the measurement of vertically opposite angles is equal

AB and CD are straight lines intersecting at O. As a result two pairs of vertically opposite angles are formed- ∠AOD, ∠BOC, and ∠AOC and ∠BOD. Prove that ∠AOD=∠BOC and ∠AOC= ZBOD.

Proof: Straight-line CD stands on OA

∴ ∠AOD + ∠AOC = 180°

Again, AB stands on a straight line OC.

∠AOC + ∠BOC = 180°

∴ ∠AOD + ∠AOC = ∠AOC + ∠BOC

Hence, ∠A0D= ∠B0C( ZSubtracting AOC from both sides) Similarly, it can be proved that ∠AOC =∠BOD.

Question 6. Find the value of ∠BOD, ∠BOC, and ∠AOC

Solution:

∴ OD stands on a straight line AB

∠AOD + ∠BOD = 180° .

or, 120° + ∠BOD = 180°

or, ∠BOD = 180° – 120° = 60°

∠QOR and ∠POR

∠BOC = Vertically Opposite Angle ∠AOD = 120°

∠AOC = Vertically Opposite Angle ∠BOD = 60°

“Class 8 WBBSE Maths Chapter 7 solutions, Vertically Opposite Angles explained”

Question 7. The sum of ∠POR and ∠QOS is 110°, find the value of ∠POS, ∠QOS,

Solution:

Given

The sum of ∠POR and ∠QOS is 110°

The sum of ∠POR and ∠QOS is 110°.

∠POR = VOA ∠QOS = \(\frac{110^{\circ}}{2}\) =55°

∠POS = 180° – 55° = 125°

∠QOR = VOA ∠POS = 125°

Question 8. OP, OQ, OR, and OS are concurrent. OP and OR are on the same straight line. P and R point are situated on opposite sides of O. ∠POQ = Z ROS and ∠POQ = ∠QOR. If ∠POQ = 50° then find the measurement of ∠QOR, ∠ROS, and ∠POS.

Solution:

Given

OP, OQ, OR, and OS are concurrent. OP and OR are on the same straight line. P and R point are situated on opposite sides of O. ∠POQ = Z ROS and ∠POQ = ∠QOR. If ∠POQ = 50°

∠POQ + ∠QOR = 180°

or, 50° + ∠QOR = 180°

or, ∠QOR = 180° – 50° – 130°

∠ROS = VOA ∠POQ = 50°

∠POS = VOA ∠QOR = 130°

Question 9. Four rays meet at a point in such a way that the measurement of opposite angles is equal. Let’s prove that two straight lines are formed by those four rays.

Solution:

Given

Four rays meet at a point in such a way that the measurement of opposite angles are equal.

Let OA, OB, OC, and OD four rays meet at the O point, as a result of which ∠AOB = Vertically Opposite Angle ∠COD and ∠AOD = Vertically Opposite Angle ∠BOC. We have to prove OA, OC and BO, OD lie on the same straight line.

“WBBSE Class 8 Maths Chapter 7, Concept of Vertically Opposite Angles solved examples”

Proof : ∠AOB + ∠AOD = ∠COD + ∠BOC

= 1/2 (∠AOB + ∠AOD + ∠COD + ∠BOC)

= \(\frac{1}{2}\)x360°

= 180°

∴  ∠AOB and ∠AOD are two adjacent angles whose sum of measurements is equal to two right angles.

∴ Their external sides OB and OD lie on the same straight line.

Similarly, it can be proved that OA and OC lie on the same straight line.

Question 10. Let’s prove that the internal and external bisectors of an angle are perpendicular to each other.

Solution:

Let the external  ∠BOC be ∠AOC OD be the internal bisector of ∠BOC and OE be the bisector of ∠AOC.

We have to prove that ∠COD + ∠COE = 1 right angle.

Proof: OC stands on a straight line AB ∠BOC + ∠AOC = 2 right angles 2 ∠COD + 2 ∠COE = 2 right angles or, ∠COD+ ∠COE = 1 right angle or, ∠EOD = 1 right angle So, OD and OE are perpendicular to each other.

∴ The internal and external bisectors of an angle are perpendicular to each other.

“WBBSE Class 8 Vertically Opposite Angles solutions, Maths Chapter 7”

Question 11. If two straight lines intersect each other then four angles are formed. Let’s prove that the sum of measures of the four angles is four right angles.

Solution:

Given

If two straight lines intersect each other then four angles are formed.

Straight lines AB and CD intersect at O

Prove that ∠AOD + ∠DOB + ∠BOC + ∠COA = 4 right angles.

Proof: OD stands on a straight line AB

∠AOD + ∠DOB = 2 right angles

Again, AB straight line stands on OC

∴ ∠BOC + ∠COA = 2 right angles

∠AOD + ∠DBO + ∠BOC + ∠COA = 4 right angles

“Class 8 WBBSE Maths Chapter 7, Vertically Opposite Angles easy explanation”

Question 12. In triangle PQR, Z PQR = z PRQ. If we extend QR on both sides then two exterior angles are formed. Let’s prove that the measurement of external angles are formed. Let’s prove that the bisectors of these angles are two perpendicular straight lines.

Solution:

Given

In triangle PQR, Z PQR = z PRQ. If we extend QR on both sides then two exterior angles are formed.

In triangle PQR ∠PQR = ∠PRQ. On both sides of QR, C, and D are extended. We have to prove that ∠PQC = ∠PRD.

Proof: PQ stands on CD straight line ∠PQC + ∠PQR = 180°

Hence, straight-line PR stands on the straight-line CD.

∴ ∠PRQ + ∠PRD = 180°

∴ ∠PQC + ∠PQR = ∠PRQ + ∠PRD

∴ ∠PQR – ∠PRQ (Given)

∠PQC = ∠PRD Proved

“WBBSE Class 8 Maths Chapter 7 solutions, Vertically Opposite Angles PDF”

Question 13. Two straight lines intersect each other at a point and thus four angles are formed. Let’s prove that the bisectors of these angles are two perpendicular straight lines.

Solution:

Given

Two straight lines intersect each other at a point and thus four angles are formed.

Let two straight lines AB and CD intersect at O. AOD, ∠DOB, ∠BOC, and ∠AOC are formed. Again let OF, OE, OH, and OG are respectively the external bisectors of ∠AOD, ∠DOB, ∠BOC, and ∠AOC. We have to prove that OE⊥LOF, OG⊥LOH, OF⊥ LOG, and OH⊥OE

“WBBSE Class 8 Maths Chapter 7, Vertically Opposite Angles important questions”

Proof:  OD stands on a straight line AB

∠AOD + ∠BOD = 2 right angles

or, \(\frac{1}{2} \angle \mathrm{AOD}+\frac{1}{2} \angle \mathrm{BOD}=\frac{1}{2} \times 2\) right angles

or, ∠DOF + ∠DOE = 1 right angle

or, ∠EOF = 1 right angle

∴ OE ⊥OF

Similarly, it can proved that OG⊥LOH, OF ⊥OG, and OH⊥OE Proved.