Chapter 14: Lines and Angles Exercise – 14.1

Question: 1Write down each pair of adjacent angles shown in Figure

Solution:The angles that have common vertex and a common arm are known as adjacent angles

The adjacent angles are:

∠DOC and ∠BOC

∠COB and ∠BOA

Question: 2In figure, name all the pairs of adjacent angles.

Solution:In fig (i), the adjacent angles are

∠EBA and ∠ABC

∠ACB and ∠BCF

∠BAC and ∠CAD

In fig (ii), the adjacent angles are

∠BAD and ∠DAC

∠BDA and ∠CDA

Question: 3In fig , write down

(i) each linear pair

(ii) each pair of vertically opposite angles.

Solution:(i) The two adjacent angles are said to form a linear pair of angles if their non – common arms are two opposite rays.

∠1 and ∠3

∠1 and ∠2

∠4 and ∠3

∠4 and ∠2

∠5 and ∠6

∠5 and ∠7

∠6 and ∠8

∠7 and ∠8

(ii) The two angles formed by two intersecting lines and have no common arms are called vertically opposite angles.

∠1 and ∠4

∠2 and ∠3

∠5 and ∠8

∠6 and ∠7

Question: 4Are the angles 1 and 2 in figure adjacent angles?

Solution:No, because they do not have common vertex.

Question: 5Find the complement of each of the following angles:

(i) 35°

(ii) 72°

(iii) 45°

(iv) 85°

Solution:The two angles are said to be complementary angles if the sum of those angles is 90°

Complementary angles for the following angles are:

(i) 90° – 35° = 55°

(ii) 90° – 72° = 18°

(iii) 90° – 45° = 45°

(iv) 90° – 85° = 5°

Question: 6Find the supplement of each of the following angles:

(i) 70°

(ii) 120°

(iii) 135°

(iv) 90°

Solution:The two angles are said to be supplementary angles if the sum of those angles is 180°

(i) 180° – 70° = 110°

(ii) 180° – 120° = 60°

(iii) 180° – 135° = 45°

(iv) 180° – 90° = 90°

Question: 7Identify the complementary and supplementary pairs of angles from the following pairs

(i) 25°, 65°

(ii) 120°, 60°

(iii) 63°, 27°

(iv) 100°, 80°

Solution:(i) 25° + 65° = 90° so, this is a complementary pair of angle.

(ii) 120° + 60° = 180° so, this is a supplementary pair of angle.

(iii) 63° + 27° = 90° so, this is a complementary pair of angle.

(iv) 100° + 80° = 180° so, this is a supplementary pair of angle.

Here, (i) and (iii) are complementary pair of angles and (ii) and (iv) are supplementary pair of angles.

Question: 8Can two obtuse angles be supplementary, if both of them be

(i) obtuse?

(ii) right?

(iii) acute?

Solution:(i) No, two obtuse angles cannot be supplementary

Because, the sum of two angles is greater than 90 degrees so their sum will be greater than 180degrees.

(ii) Yes, two right angles can be supplementary

Because, 90° + 90° = 180°

(iii) No, two acute angle cannot be supplementary

Because, the sum of two angles is less than 90 degrees so their sum will also be less tha 90 degrees.

Question: 9Name the four pairs of supplementary angles shown in Fig.

Solution:The supplementary angles are

∠AOC and ∠COB

∠BOC and ∠DOB

∠BOD and ∠DOA

∠AOC and ∠DOA

Question: 10In Figure, A, B, C are collinear points and ∠DBA = ∠EBA.

(i) Name two linear pairs.

(ii) Name two pairs of supplementary angles.

Solution:(i) Linear pairs

∠ABD and ∠DBC

∠ABE and ∠EBC

Because every linear pair forms supplementary angles, these angles are

∠ABD and ∠DBC

∠ABE and ∠EBC

Question: 11If two supplementary angles have equal measure, what is the measure of each angle?

Solution:Let p and q be the two supplementary angles that are equal

∠p = ∠q

So,

∠p + ∠q = 180°

=> ∠p + ∠p = 180°

=> 2∠p = 180°

=> ∠p = 180°/2

=> ∠p = 90°

Therefore, ∠p = ∠q = 90°

Question: 12If the complement of an angle is 28°, then find the supplement of the angle.

Solution:Here, let p be the complement of the given angle 28°

Therefore, ∠p + 28° = 90°

=> ∠p = 90° – 28°

= 62°

So, the supplement of the angle = 180° – 62°

= 118°

Question: 13In Figure, name each linear pair and each pair of vertically opposite angles.

Solution:Two adjacent angles are said to be linear pair of angles, if their non-common arms are two opposite rays.

∠1 and ∠2

∠2 and ∠3

∠3 and ∠4

∠1 and ∠4

∠5 and ∠6

∠6 and ∠7

∠7 and ∠8

∠8 and ∠5

∠9 and ∠10

∠10 and ∠11

∠11 and ∠12

∠12 and ∠9

The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.

∠1 and ∠3

∠4 and ∠2

∠5 and ∠7

∠6 and ∠8

∠9 and ∠11

∠10 and ∠12

Question: 14In Figure, OE is the bisector of ∠BOD. If ∠1 = 70°, Find the magnitude of ∠2, ∠3, ∠4

Solution:Given,

∠1 = 70°

∠3 = 2(∠1)

= 2(70°)

= 140°

∠3 = ∠4

As, OE is the angle bisector,

∠DOB = 2(∠1)

= 2(70°)

= 140°

∠DOB + ∠AOC + ∠COB +∠DOB = 360°

=> 140° + 140° + 2(∠COB) = 360°

Since, ∠COB = ∠AOD

=> 2(∠COB) = 360° – 280°

=> 2(∠COB) = 80°

=> ∠COB = 80°/2

=> ∠COB = 40°

Therefore, ∠COB = ∠AOB = 40°

The angles are,

∠1 = 70°,

∠2 = 40°,

∠3 = 140°,

∠4 = 40°

Question: 15One of the angles forming a linear pair is a right angle. What can you say about its other angle?

Solution:One of the Angle of a linear pair is the right angle (90°)

Therefore, the other angle is

=> 180° – 90° = 90°

Question: 16One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

Solution:One of the Angles of a linear pair is obtuse, then the other angle should be acute, only then their sum will be 180°.

Question: 17One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

Solution:One of the Angles of a linear pair is acute, then the other angle should be obtuse, only then their sum will be 180°.

Question: 18Can two acute angles form a linear pair?

Solution:No, two acute angles cannot form a linear pair because their sum is always less than 180°.

Question: 19If the supplement of an angle is 65°, then find its complement.

Solution:Let x be the required angle

So,

=> x + 65° = 180°

=> x = 180° – 65°

= 115°

But the complement of the angle cannot be determined.

Question: 20Find the value of x in each of the following figures

**Solution:**

(i) Since, ∠BOA + ∠BOC = 180°

Linear pair:

=> 60° + x° = 180°

=> x° = 180° – 60°

=> x° = 120°

(ii)

=> 3x° + 2x° = 180°

=> 5x° = 180°

=> x° = 180°/5

=> x° = 36°

(iii)

Since, 35° + x° + 60° = 180°

=> x° = 180° – 35° – 60°

=> x° = 180° – 95°

=> x° = 85°

(iv) Linear pair,

83° + 92° + 47° + 75° + x° = 360°

=> x° + 297° = 360°

=> x° = 360° – 297°

=> x° = 63°

(v) Linear pair,

3x° + 2x° + x° + 2x° = 360°

=> 8x° = 360°

=> x° = 360°/8

=> x∘ = 45°

(vi) Linear pair:

3x° = 105°

=> x° = 105°/3

=> x° = 45°

**Question:**

In Fig. 22, it being given that ∠1 = 65°, find all the other angles.

**Solution:**

Given,

∠1 = ∠3 are the vertically opposite angles

Therefore, ∠3 = 65°

Here, ∠1 + ∠2 = 180° are the linear pair

Therefore, ∠2 = 180° – 65°

= 115°

∠2 = ∠4 are the vertically opposite angles

Therefore, ∠2 = ∠4 = 115°

And ∠3 = 65°

**Question: 22**

In Fig. 23 OA and OB are the opposite rays:

(i) If x = 25°, what is the value of y?

(ii) If y = 35°, what is the value of x?

**Solution:**

∠AOC + ∠BOC = 180° – Linear pair

=> 2y + 5 + 3x = 180°

=> 3x + 2y = 175°

(i) If x = 25°, then

=> 3(25°) + 2y = 175°

=> 75° + 2y = 175°

=> 2y = 175° – 75°

=> 2y = 100°

=> y = 100°/2

=> y = 50°

(ii) If y = 35°, then

3x + 2(35°) = 175°

=> 3x + 70° = 175°

=> 3x = 175° – 70°

=> 3x = 105°

=> x = 105°/3

=> x = 35°

**Question: 24**

In Figure, write all pairs of adjacent angles and all the linear pairs.

**Solution:**

Pairs of adjacent angles are:

∠DOA and ∠DOC

∠BOC and ∠COD

∠AOD and ∠BOD

∠AOC and ∠BOC

Linear pairs:

∠AOD and ∠BOD

∠AOC and ∠BOC

**Question: 25**

In Figure, find ∠x. Further find ∠BOC, ∠COD, ∠AOD

**Solution:**

(x + 10)° + x° + (x + 20) ° = 180°

=> 3x° + 30° = 180°

=> 3x° = 180° – 30°

=> 3x° = 150°

=> x° = 150°/3

=> x° = 50°

Here,

∠BOC = (x + 20)°

= (50 + 20)°

= 70°

∠COD = 50°

∠AOD = (x + 10)°

= (50 + 10)°

= 60°

**Question: 25**

How many pairs of adjacent angles are formed when two lines intersect in a point?

**Solution:**

If the two lines intersect at a point, then four adjacent pairs are formed and those are linear.

**Question: 26**

How many pairs of adjacent angles, in all, can you name in Figure?

**Solution:**

There are 10 adjacent pairs

∠EOD and ∠DOC

∠COD and ∠BOC

∠COB and ∠BOA

∠AOB and ∠BOD

∠BOC and ∠COE

∠COD and ∠COA

∠DOE and ∠DOB

∠EOD and ∠DOA

∠EOC and ∠AOC

∠AOB and ∠BOE

**Question: 27**

In Figure, determine the value of x.

**Solution:**

Linear pair:

∠COB + ∠AOB = 180°

=> 3x° + 3x° = 180°

=> 6x° = 180°

=> x° = 180°/6

=> x° = 30°

**Question: 28**

In Figure, AOC is a line, find x.

**Solution:**

∠AOB + ∠BOC = 180°

Linear pair

=> 2x + 70° = 180°

=> 2x = 180° – 70°

=> 2x = 110°

=> x = 110°/2

=> x = 55°

**Question: 29**

In Figure, POS is a line, find x.

**Solution:**

Angles of a straight line,

∠QOP + ∠QOR + ∠ROS = 108°

=> 60° + 4x + 40° = 180°

=> 100° + 4x = 180°

=> 4x = 180° – 100°

=> 4x = 80°

=> x = 80°/4

=> x = 20°

**Question: 30**

In Figure, lines l_{1} and l_{2} intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.

**Solution:**

Given that,

∠x = 45°

∠x = ∠z = 45°

∠y = ∠u

∠x + ∠y + ∠z + ∠u = 360°

=> 45° + 45° + ∠y + ∠u = 360°

=> 90° + ∠y + ∠u = 360°

=> ∠y + ∠u = 360° – 90°

=> ∠y + ∠u = 270°

=> ∠y + ∠z = 270°

=> 2∠z = 270°

=> ∠z = 135°

Therefore, ∠y = ∠u = 135°

So, ∠x = 45°,

∠y = 135°,

∠z = 45°,

∠u = 135°

**Question: 31**

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

**Solution:**

Given that,

∠x + ∠y + ∠z + ∠u + 50° + 90° = 360°

Linear pair,

∠x + 50° + 90° = 180°

=> ∠x + 140° = 180°

=> ∠x = 180° – 140°

=> ∠x = 40°

∠x = ∠u = 40° are vertically opposite angles

=> ∠z = 90° is a vertically opposite angle

=> ∠y = 50° is a vertically opposite angle

Therefore, ∠x = 40°,

∠y = 50°,

∠z = 90°,

∠u = 40°

**Question: 32**

In Figure, find the values of x, y and z

**Solution:**

∠y = 25° vertically opposite angle

∠x = ∠y are vertically opposite angles

∠x + ∠y + ∠z + 25° = 360°

=> ∠x + ∠z + 25° + 25° = 360°

=> ∠x + ∠z + 50° = 360°

=> ∠x + ∠z = 360° – 50°

=> 2∠x = 310°

=> ∠x = 155°

And, ∠x = ∠z = 155°

Therefore, ∠x = 155°,

∠y = 25°,

∠z = 155°