RD Chapter 8- Lines and Angles Ex-8.1 |
RD Chapter 8- Lines and Angles Ex-8.2 |
RD Chapter 8- Lines and Angles Ex-8.4 |
RD Chapter 8- Lines and Angles Ex-VSAQS |

**Answer
1** :

Given: x = 45^{0}

Since vertically opposite angles are equal, therefore z = x = 45^{0}

z and u are angles that are a linear pair, therefore, z + u =180^{0}

Solve, z + u = 180^{0} , for u

u = 180^{0} – z

u = 180^{0} – 45

u = 135^{0}

Again, x and y angles are a linear pair.

x+ y = 180^{0}

y = 180^{0} – x

y =180^{0} – 45^{0}

y = 135^{0}

Hence, remaining angles are y = 135^{0}, u = 135^{0} andz = 45^{0}.

In figure, threecoplanar lines intersect at a point O, forming angles as shown in the figure.Find the values of x, y, z and u .

**Answer
2** :

(∠BOD,z); (∠DOF, y) are pair of vertically opposite angles.

So, ∠BOD = z= 90^{0}

∠DOF = y= 50^{0}

[Vertically oppositeangles are equal.]

Now, x + y + z = 180 [Linear pair] [AB is a straight line]

x + y + z = 180

x + 50 + 90 = 180

x = 180 – 140

x = 40

Hence values of x, y, z and u are 40^{0}, 50^{0},90^{0} and 40^{0} respectively.

**Answer
3** :

From figure,

y = 25^{0} [Vertically opposite angles are equal]

Now ∠x + ∠y = 180^{0} [Linear pair ofangles]

x = 180 – 25

x = 155

Also, z = x = 155 [Vertically opposite angles]

Answer: y = 25^{0} and z = 155^{0}

In figure, find thevalue of x.

**Answer
4** :

∠AOE = ∠BOF = 5x [Vertically opposite angles]

∠COA+∠AOE+∠EOD = 180^{0} [Linear pair]

3x + 5x + 2x = 180

10x = 180

x = 180/10

x = 18

The value of x = 18^{0}

Prove thatbisectors of a pair of vertically opposite angles are in the same straightline.

**Answer
5** :

Lines AB and CD intersect at point O, such that

∠AOC = ∠BOD (vertically angles) …(1)

Also OP is the bisector of AOC and OQ is the bisector of BOD

To Prove: POQ is a straight line.

OP is the bisector of ∠AOC:

∠AOP = ∠COP …(2)

OQ is the bisector of ∠BOD:

∠BOQ = ∠QOD …(3)

Now,

Sum of the angles around a point is 360^{o}.

∠AOC + ∠BOD + ∠AOP + ∠COP + ∠BOQ + ∠QOD = 360^{0}

∠BOQ + ∠QOD + ∠DOA + ∠AOP + ∠POC + ∠COB = 360^{0}

2∠QOD + 2∠DOA + 2∠AOP = 360^{0} (Using (1), (2) and (3))

∠QOD + ∠DOA + ∠AOP = 180^{0}

POQ = 180^{0}

Which shows that, the bisectors of pair of vertically oppositeangles are on the same straight line.

Hence Proved.

If two straightlines intersect each other, prove that the ray opposite to the bisector of oneof the angles thus formed bisects the vertically opposite angle.

**Answer
6** :

Given AB and CD are straight lines which intersect at O.

OP is the bisector of ∠ AOC.

To Prove : OQ is the bisector of ∠BOD

Proof :

AB, CD and PQ are straight lines which intersect in O.

Vertically opposite angles: ∠ AOP = ∠ BOQ

Vertically opposite angles: ∠ COP = ∠ DOQ

OP is the bisector of ∠ AOC : ∠ AOP= ∠ COP

Therefore, ∠BOQ = ∠ DOQ

Hence, OQ is the bisector of ∠BOD.

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