# RD Sharma Solutions for Class 7 Maths Chapter 14 Lines And Angles

Get free PDF of RD Sharma Solutions for Class 7 Maths Chapter 14 Lines and Angles from the given links. Students who aim to top the examinations can refer to RD Sharma Solutions. BYJUâ€™S expert team has solved the questions in such a way that learners can understand easily and comfortably. Learners can simply download the PDFs from the available links.

Chapter 14, Lines And Angles includes two exercises. RD Sharma Solutions for Class 7 includes answers to all questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this Chapter.

• Pairs of Angles
• Definition and meaning of adjacent angles
• Definition and meaning of linear pair
• Vertically opposite angles
• Definition of angles at a point
• Concept of complementary angles
• Concept of supplementary angles
• Definition and meaning of parallel lines
• Parallel rays
• Parallel segments
• Transversals
• Angle made by a transversal with two lines
• Exterior and interior angles
• Corresponding angles
• Alternate interior and interior angles
• Angles made by a transversal to two parallel lines

## Download the PDF of RD Sharma Solutions For Class 7 Maths Chapter 14 Lines And Angles

### Access answers to Maths RD Sharma Solutions For Class 7 Chapter 14 – Lines And Angles

Exercise 14.1 Page No: 14.6

1. Write down each pair of adjacent angles shown in fig. 13.

Solution:

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

Therefore the adjacent angles in given figure are:

âˆ DOC and âˆ BOC

âˆ COB and âˆ BOA

2. In Fig. 14, name all the pairs of adjacent angles.

Solution:

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

In fig (i), the adjacent angles are

âˆ EBA and âˆ ABC

âˆ ACB and âˆ BCF

In fig (ii), the adjacent angles are

âˆ BDA and âˆ CDA

3. In fig. 15, 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 â€“ common arms are two opposite rays.

âˆ 1 and âˆ 3

âˆ 1 and âˆ 2

âˆ 4 and âˆ 3

âˆ 4 and âˆ 2

âˆ 5 and âˆ 6

âˆ 5 and âˆ 7

âˆ 6 and âˆ 8

âˆ 7 and âˆ 8

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

âˆ 1 and âˆ 4

âˆ 2 and âˆ 3

âˆ 5 and âˆ 8

âˆ 6 and âˆ 7

4. Are the angles 1 and 2 given in Fig. 16 adjacent angles?

Solution:

No, because they donâ€™t have common vertex.

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

(i) 35o

(ii) 72o

(iii) 45o

(iv) 85o

Solution:

(i) The two angles are said to be complementary angles if the sum of those angles is 90o

Complementary angle for given angle is

90o â€“ 35o = 55o

(ii) The two angles are said to be complementary angles if the sum of those angles is 90o

Complementary angle for given angle is

900 â€“ 72o = 18o

(iii) The two angles are said to be complementary angles if the sum of those angles is 90o

Complementary angle for given angle is

90o â€“ 45o = 45o

(iv) The two angles are said to be complementary angles if the sum of those angles is 90o

Complementary angle for given angle is

90o â€“ 85o = 5o

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

(i) 70o

(ii) 120o

(iii) 135o

(iv) 90o

Solution:

(i) The two angles are said to be supplementary angles if the sum of those angles is 180o

Therefore supplementary angle for the given angle is

180o â€“ 70o = 110o

(ii) The two angles are said to be supplementary angles if the sum of those angles is 180o

Therefore supplementary angle for the given angle is

180o â€“ 120o = 60o

(iii) The two angles are said to be supplementary angles if the sum of those angles is 180o

Therefore supplementary angle for the given angle is

180o â€“ 135o = 45o

(iv) The two angles are said to be supplementary angles if the sum of those angles is 180o

Therefore supplementary angle for the given angle is

180o â€“ 90o = 90o

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

(i) 25o, 65o

(ii) 120o, 60o

(iii) 63o, 27o

(iv) 100o, 80o

Solution:

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

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

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

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

8. Can 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 90o so their sum will be greater than 180o

(ii) Yes, two right angles can be supplementary

Because, 90o + 90o = 180o

(iii) No, two acute angle cannot be supplementary

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

9. Name the four pairs of supplementary angles shown in Fig.17.

Solution:

The two angles are said to be supplementary angles if the sum of those angles is 180o

The supplementary angles are

âˆ AOC and âˆ COB

âˆ BOC and âˆ DOB

âˆ BOD and âˆ DOA

âˆ AOC and âˆ DOA

10. In Fig. 18, A, B, C are collinear points and âˆ DBA = âˆ EBA.

(i) Name two linear pairs.

(ii) Name two pairs of supplementary angles.

Solution:

(i) Two adjacent angles are said to be form a linear pair of angles, if their non-common arms are two opposite rays.

Therefore linear pairs are

âˆ ABD and âˆ DBC

âˆ ABE and âˆ EBC

(ii) We know that every linear pair forms supplementary angles, these angles are

âˆ ABD and âˆ DBC

âˆ ABE and âˆ EBC

11. If 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

The two angles are said to be supplementary angles if the sum of those angles is 180o

âˆ p = âˆ q

So,

âˆ p + âˆ q = 180o

âˆ p + âˆ p = 180o

2âˆ p = 180o

âˆ p = 180o/2

âˆ p = 90o

Therefore, âˆ p = âˆ q = 90o

12. If the complement of an angle is 28o, then find the supplement of the angle.

Solution:

Given complement of an angle is 28o

Here, let x be the complement of the given angle 28o

Therefore, âˆ x + 28o = 90o

âˆ x = 90o â€“ 28o

= 62o

So, the supplement of the angle = 180o â€“ 62o

= 118o

13. In Fig. 19, 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.

Therefore linear pairs are listed below:

âˆ 1 and âˆ 2

âˆ 2 and âˆ 3

âˆ 3 and âˆ 4

âˆ 1 and âˆ 4

âˆ 5 and âˆ 6

âˆ 6 and âˆ 7

âˆ 7 and âˆ 8

âˆ 8 and âˆ 5

âˆ 9 and âˆ 10

âˆ 10 and âˆ 11

âˆ 11 and âˆ 12

âˆ 12 and âˆ 9

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

Therefore supplement of the angle are listed below:

âˆ 1 and âˆ 3

âˆ 4 and âˆ 2

âˆ 5 and âˆ 7

âˆ 6 and âˆ 8

âˆ 9 and âˆ 11

âˆ 10 and âˆ 12

14. In Fig. 20, OE is the bisector of âˆ BOD. If âˆ 1 = 70o, find the magnitude of âˆ 2, âˆ 3 and âˆ 4.

Solution:

Given, âˆ 1 = 70o

âˆ 3 = 2(âˆ 1)

= 2(70o)

âˆ 3 = 140o

âˆ 3 = âˆ 4

As, OE is the angle bisector,

âˆ DOB = 2(âˆ 1)

= 2(70o)

= 140o

âˆ DOB + âˆ AOC + âˆ COB +âˆ AOD = 360o [sum of the angle of circle = 360o]

140o + 140o + 2(âˆ COB) = 360o

Since, âˆ COB = âˆ AOD

2(âˆ COB) = 360o â€“ 280o

2(âˆ COB) = 80o

âˆ COB = 80o/2

âˆ COB = 40o

Therefore, âˆ COB = âˆ AOD = 40o

The angles are, âˆ 1 = 70o, âˆ 2 = 40o, âˆ 3 = 140o and âˆ 4 = 40o

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

Solution:

Given one of the angle of a linear pair is the right angle that is 90o

We know that linear pair angle is 180o

Therefore, the other angle is

180o â€“ 90o = 90o

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

Solution:

Given one of the angles of a linear pair is obtuse, then the other angle should be acute, because only then their sum will be 180o.

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

Solution:

Given one of the Angles of a linear pair is acute, then the other angle should be obtuse, only then their sum will be 180o.

18. Can two acute angles form a linear pair?

Solution:

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

19. If the supplement of an angle is 65o, then find its complement.

Solution:

Let x be the required angle

So, x + 65o = 180o

x = 180o â€“ 65o

x = 115o

The two angles are said to be complementary angles if the sum of those angles is 90o here it is more than 90o therefore the complement of the angle cannot be determined.

20. Find the value of x in each of the following figures.

Solution:

(i) Â We know that âˆ BOA + âˆ BOC = 180o

[Linear pair: The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

60o + xo = 180o

xo = 180o â€“ 60o

xo = 120o

(ii) We know that âˆ POQ + âˆ QOR = 180o

[Linear pair: The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

3xo + 2xo = 180o

5xo = 180o

xo = 180o/5

xo = 36o

(iii) We know that âˆ LOP + âˆ PON + âˆ NOM = 180o

[Linear pair: The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

Since, 35o + xo + 60o = 180o

xo = 180o â€“ 35o â€“ 60o

xo = 180o â€“ 95o

xo = 85o

(iv) We know that âˆ DOC + âˆ DOE + âˆ EOA + âˆ AOB+ âˆ BOC = 360o

83o + 92o + 47o + 75o + xo = 360o

xo + 297o = 360o

xo = 360o â€“ 297o

xo = 63o

(v) We know that âˆ ROS + âˆ ROQ + âˆ QOP + âˆ POS = 360o

3xo + 2xo + xo + 2xo = 360o

8xo = 360o

xo = 360o/8

xo = 45o

(vi) Â Linear pair: The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o

Therefore 3xo = 105o

xo = 105o/3

xo = 35o

21. In Fig. 22, it being given that âˆ 1 = 65o, find all other angles.

Solution:

Given from the figure 22, âˆ 1 = âˆ 3 are the vertically opposite angles

Therefore, âˆ 3 = 65o

Here, âˆ 1 + âˆ 2 = 180Â° are the linear pair [The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

Therefore, âˆ 2 = 180o â€“ 65o

= 115o

âˆ 2 = âˆ 4 are the vertically opposite angles [from the figure]

Therefore, âˆ 2 = âˆ 4 = 115o

And âˆ 3 = 65o

22. In Fig. 23, OA and OB are opposite rays:

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

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

Solution:

(i) âˆ AOC + âˆ BOC = 180o [The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

2y + 50 + 3x = 180o

3x + 2y = 175o

Given If x = 25o, then

3(25o) + 2y = 175o

75o + 2y = 175o

2y = 175o â€“ 75o

2y = 100o

y = 100o/2

y = 50o

(ii) âˆ AOC + âˆ BOC = 180o [The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

2y + 5 + 3x = 180o

3x + 2y = 175o

Given If y = 35o, then

3x + 2(35o) = 175o

3x + 70o = 175o

3x = 1750 â€“ 70o

3x = 105o

x = 105o/3

x = 35o

23. In Fig. 24, write all pairs of adjacent angles and all the liner pairs.

Solution:

âˆ DOA and âˆ DOC

âˆ BOC and âˆ COD

âˆ AOD and âˆ BOD

âˆ AOC and âˆ BOC

Linear pairs: [The two adjacent angles are said to form a linear pair of angles if their nonâ€“common arms are two opposite rays and sum of the angle is 180o]

âˆ AOD and âˆ BOD

âˆ AOC and âˆ BOC

24. In Fig. 25, find âˆ x. Further find âˆ BOC, âˆ COD and âˆ AOD.

Solution:

(x + 10)o + xo + (x + 20)o = 180o[linear pair]

On rearranging we get

3xo + 30o = 180o

3xo = 180o â€“ 30o

3xo = 150o

xo = 150o/3

xo = 50o

Also given that

âˆ BOC = (x + 20)o

= (50 + 20)o

= 70o

âˆ COD = 50o

âˆ AOD = (x + 10)o

= (50 + 10)o

= 60o

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.

26. How many pairs of adjacent angles, in all, can you name in Fig. 26?

Solution:

There are 10 adjacent pairs formed in the given figure, they are

âˆ EOD and âˆ DOC

âˆ COD and âˆ BOC

âˆ COB and âˆ BOA

âˆ AOB and âˆ BOD

âˆ BOC and âˆ COE

âˆ COD and âˆ COA

âˆ DOE and âˆ DOB

âˆ EOD and âˆ DOA

âˆ EOC and âˆ AOC

âˆ AOB and âˆ BOE

27. In Fig. 27, determine the value of x.

Solution:

From the figure we can write as âˆ COB + âˆ AOB = 180o [linear pair]

3xo + 3xo = 180o

6xo = 180o

xo = 180o/6

xo = 30o

28. In Fig.28, AOC is a line, find x.

Solution:

From the figure we can write as

âˆ AOB + âˆ BOC = 180o [linear pair]

Linear pair

2x + 70o = 180o

2x = 180o â€“ 70o

2x = 110o

x = 110o/2

x = 55o

29. In Fig. 29, POS is a line, find x.

Solution:

From the figure we can write as angles of a straight line,

âˆ QOP + âˆ QOR + âˆ ROS = 180o

60o + 4x + 40o = 180o

On rearranging we get, 100o + 4x = 180o

4x = 180o â€“ 100o

4x = 80o

x = 80o/4

x = 20o

30. In Fig. 30, lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45o, find the values of y, z and u.

Solution:

Given that, âˆ x = 45o

From the figure we can write as

âˆ x = âˆ z = 45o

Also from the figure, we have

âˆ y = âˆ u

From the property of linear pair we can write as

âˆ x + âˆ y + âˆ z + âˆ u = 360o

45o + 45o + âˆ y + âˆ u = 360o

90o + âˆ y + âˆ u = 360o

âˆ y + âˆ u = 360o â€“ 90o

âˆ y + âˆ u = 270o (vertically opposite angles âˆ y = âˆ u)

2âˆ y = 270o

âˆ y = 135o

Therefore, âˆ y = âˆ u = 135o

So, âˆ x = 45o, âˆ y = 135o, âˆ z = 45o and âˆ u = 135o

31. In Fig. 31, 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, âˆ x + âˆ y + âˆ z+ âˆ u + 50o + 90o = 360o

Linear pair, âˆ x + 50o + 90o = 180o

âˆ x + 140o = 180o

On rearranging we get

âˆ x = 180o â€“ 140o

âˆ x = 40o

From the figure we can write as

âˆ x = âˆ u = 40o are vertically opposite angles

âˆ z = 90o is a vertically opposite angle

âˆ y = 50o is a vertically opposite angle

Therefore, âˆ x = 40o, âˆ y = 50o, âˆ z = 90o and âˆ u = 40o

32. In Fig. 32, find the values of x, y and z.

Solution:

âˆ y = 25o vertically opposite angle

From the figure we can write as

âˆ x = âˆ z are vertically opposite angles

âˆ x + âˆ y + âˆ z + 25o = 360o

âˆ x + âˆ z + 25o + 25o = 360o

On rearranging we get,

âˆ x + âˆ z + 50o = 360o

âˆ x + âˆ z = 360o â€“ 50o [âˆ x = âˆ z]

2âˆ x = 310o

âˆ x = 155o

And, âˆ x = âˆ z = 155o

Therefore, âˆ x = 155o, âˆ y = 25o and âˆ z = 155o

Exercise 14.2 Page No: 14.20

1. In Fig. 58,Â line n is a transversal to line l and m. Identify the following:

(i) Alternate and corresponding angles in Fig. 58 (i)

(ii) Angles alternate to âˆ d and âˆ g and angles corresponding to âˆ f and âˆ h in Fig. 58 (ii)

(iii) Angle alternate to âˆ PQR, angle corresponding to âˆ RQF and angle alternate to âˆ PQE in Fig. 58 (iii)

(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. 58 (ii)

Solution:

(i) A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.

In Figure (i) Corresponding angles are

âˆ EGB andÂ âˆ GHD

âˆ HGB andÂ âˆ FHD

âˆ EGA andÂ âˆ GHC

âˆ AGH andÂ âˆ CHF

A pair of angles in which one arm of each of the angle is on opposite sides of the transversal and whose other arms include the one segment is called a pair of alternate angles.

The alternate angles are:

âˆ EGB andÂ âˆ CHF

âˆ HGB andÂ âˆ CHG

âˆ EGA andÂ âˆ FHD

âˆ AGH andÂ âˆ GHD

(ii) In Figure (ii)

The alternate angle to âˆ d is âˆ e.

The alternate angle to âˆ g is âˆ b.

The corresponding angle to âˆ f is âˆ c.

The corresponding angle to âˆ h is âˆ a.

(iii) In Figure (iii)

Angle alternate to âˆ PQR is âˆ QRA.

Angle corresponding to âˆ RQF is âˆ ARB.

Angle alternate to âˆ POE is âˆ ARB.

(iv) In Figure (ii)

Pair of interior angles are

âˆ a is âˆ e.

âˆ d is âˆ f.

Pair of exterior angles are

âˆ b is âˆ h.

âˆ c is âˆ g.

2. In Fig. 59, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively, If âˆ CMQ = 60o, find all other angles in the figure.

Solution:

A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.

Therefore corresponding angles are

âˆ ALM = âˆ CMQ = 60o [given]

Vertically opposite angles are

âˆ LMD = âˆ CMQ = 60o [given]

Vertically opposite angles are

âˆ ALM = âˆ PLB = 60o

Here, âˆ CMQ + âˆ QMD = 180o are the linear pair

On rearranging we get

âˆ QMD = 180o â€“ 60o

= 120o

Corresponding angles are

âˆ QMD = âˆ MLB = 120o

Vertically opposite angles

âˆ QMD = âˆ CML = 120o

Vertically opposite angles

âˆ MLB = âˆ ALP = 120o

3. In Fig. 60, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If âˆ LMD = 35o find âˆ ALM and âˆ PLA.

Solution:

Given that, âˆ LMD = 35o

From the figure we can write

âˆ LMD and âˆ LMC is a linear pair

âˆ LMD + âˆ LMC = 180o [sum of angles in linear pair = 180o]

On rearranging, we get

âˆ LMC = 180o â€“ 35o

= 145o

So, âˆ LMC = âˆ PLA = 145o

And, âˆ LMC = âˆ MLB = 145o

âˆ MLB and âˆ ALM is a linear pair

âˆ MLB + âˆ ALM = 180o [sum of angles in linear pair = 180o]

âˆ ALM = 180o â€“ 145o

âˆ ALM = 350

Therefore, âˆ ALM = 35o, âˆ PLA = 145o.

4. The line n is transversal to line l and m in Fig. 61. Identify the angle alternate to âˆ 13, angle corresponding to âˆ 15, and angle alternate to âˆ 15.

Solution:

Given that, l âˆ¥ m

From the figure the angle alternate to âˆ 13 is âˆ 7

From the figure the angle corresponding to âˆ 15 is âˆ 7 [A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.]

Again from the figure angle alternate to âˆ 15 is âˆ 5

5. In Fig. 62, line l âˆ¥ m and n is transversal. If âˆ 1 = 40Â°, find all the angles and check that all corresponding angles and alternate angles are equal.

Solution:

Given that, âˆ 1 = 40o

âˆ 1 and âˆ 2 is a linear pair [from the figure]

âˆ 1 + âˆ 2 = 180o

âˆ 2 = 180o â€“ 40o

âˆ 2 = 140o

Again from the figure we can say that

âˆ 2 and âˆ 6Â is a corresponding angle pair

So, âˆ 6 = 140o

âˆ 6 and âˆ 5 is a linear pair [from the figure]

âˆ 6 + âˆ 5 = 180o

âˆ 5 = 180o â€“ 140o

âˆ 5 = 40o

From the figure we can write as

âˆ 3 and âˆ 5Â are alternate interior angles

So, âˆ 5 = âˆ 3 = 40o

âˆ 3 and âˆ 4 is a linear pair

âˆ 3 + âˆ 4 = 180o

âˆ 4 = 180o â€“ 40o

âˆ 4 = 140o

Now, âˆ 4 and âˆ 6Â are a pair of interior angles

So, âˆ 4 = âˆ 6 = 140o

âˆ 3 and âˆ 7 are a pair of corresponding angles

So, âˆ 3 = âˆ 7 = 40o

Therefore, âˆ 7 = 40o

âˆ 4 and âˆ 8Â are a pair of corresponding angles

So, âˆ 4 = âˆ 8 = 140o

Therefore, âˆ 8 = 140o

Therefore, âˆ 1 = 40o, âˆ 2 = 140o, âˆ 3 = 40o, âˆ 4 = 140o, âˆ 5 = 40o, âˆ 6 = 140o, âˆ 7 = 40o and âˆ 8 = 140o

6. In Fig.63, line l âˆ¥ m and a transversal n cuts them P and Q respectively. If âˆ 1 = 75Â°, find all other angles.

Solution:

Given that, l âˆ¥ m and âˆ 1 = 75o

âˆ 1 = âˆ 3 are vertically opposite angles

We know that, from the figure

âˆ 1 + âˆ 2 = 180o is a linear pair

âˆ 2 = 180o â€“ 75o

âˆ 2 = 105o

Here, âˆ 1 = âˆ 5 = 75o are corresponding angles

âˆ 5 = âˆ 7 = 75o are vertically opposite angles.

âˆ 2 = âˆ 6 = 105o are corresponding angles

âˆ 6 = âˆ 8 = 105o are vertically opposite angles

âˆ 2 = âˆ 4 = 105o are vertically opposite angles

So, âˆ 1 = 75o, âˆ 2 = 105o, âˆ 3 = 75o, âˆ 4 = 105o, âˆ 5 = 75o, âˆ 6 = 105o, âˆ 7 = 75o and âˆ 8 = 105o

7. In Fig. 64, AB âˆ¥ CD and a transversal PQ cuts at L and M respectively. If âˆ QMD = 100o, find all the other angles.

Solution:

Given that, AB âˆ¥ CD and âˆ QMD = 100o

We know that, from the figure âˆ QMD + âˆ QMC = 180o is a linear pair,

âˆ QMC = 180o â€“ âˆ QMD

âˆ QMC = 180o â€“ 100Â°

âˆ QMC = 80o

Corresponding angles are

âˆ DMQ = âˆ BLM = 100o

âˆ CMQ = âˆ ALM = 80o

Vertically Opposite angles are

âˆ DMQ = âˆ CML = 100o

âˆ BLM = âˆ PLA = 100o

âˆ CMQ = âˆ DML = 80o

âˆ ALM = âˆ PLB = 80o

8. In Fig. 65, l âˆ¥ m and p âˆ¥ q. Find the values of x, y, z, t.

Solution:

Given that one of the angle is 80o

âˆ z and 80o are vertically opposite angles

Therefore âˆ z = 80o

âˆ z and âˆ t are corresponding angles

âˆ z = âˆ t

Therefore, âˆ t = 80o

âˆ z and âˆ y are corresponding angles

âˆ z = âˆ y

Therefore, âˆ y = 80o

âˆ x and âˆ y are corresponding angles

âˆ y = âˆ x

Therefore, âˆ x = 80o

9. In Fig. 66, line l âˆ¥ m, âˆ 1 = 120o and âˆ 2 = 100o, find out âˆ 3 and âˆ 4.

Solution:

Given that, âˆ 1 = 120o and âˆ 2 = 100o

From the figure âˆ 1 and âˆ 5 is a linear pair

âˆ 1 + âˆ 5 = 180o

âˆ 5 = 180o â€“ 120o

âˆ 5 = 60o

Therefore, âˆ 5 = 60o

âˆ 2 and âˆ 6 are corresponding angles

âˆ 2 = âˆ 6 =Â 100o

Therefore,Â âˆ 6 =Â 100o

âˆ 6 and âˆ 3 a linear pair

âˆ 6 + âˆ 3 = 180o

âˆ 3 = 180o â€“ 100o

âˆ 3 = 80o

Therefore, âˆ 3 = 80o

By, angles of sum property

âˆ 3 + âˆ 5 + âˆ 4Â = 180o

âˆ 4 = 180o â€“ 80o â€“ 60o

âˆ 4 = 40o

Therefore, âˆ 4 = 40o

10. Â In Fig. 67, l âˆ¥ m. Find the values of a, b, c, d. Give reasons.

Solution:

Given l âˆ¥ m

From the figure vertically opposite angles,

âˆ a = 110o

Corresponding angles, âˆ a = âˆ b

Therefore, âˆ b = 110o

Vertically opposite angle,

âˆ d = 85o

Corresponding angles, âˆ d = âˆ c

Therefore, âˆ c = 85o

Hence, âˆ a = 110o, âˆ b = 110o, âˆ c = 85o, âˆ d = 85o

11. In Fig. 68, AB âˆ¥ CD and âˆ 1 and âˆ 2 are in the ratio of 3: 2. Determine all angles from 1 to 8.

Solution:

Given âˆ 1 and âˆ 2 are in the ratio 3: 2

Let us take the angles as 3x, 2x

âˆ 1 and âˆ 2 are linear pair [from the figure]

3x + 2x = 180o

5x = 180o

x = 180o/5

x = 36o

Therefore, âˆ 1 = 3x = 3(36) = 108o

âˆ 2 = 2x = 2(36) = 72o

âˆ 1 and âˆ 5 are corresponding angles

Therefore âˆ 1 = âˆ 5

Hence, âˆ 5 = 108o

âˆ 2 and âˆ 6 are corresponding angles

So âˆ 2 = âˆ 6

Therefore, âˆ 6 = 72o

âˆ 4 and âˆ 6 are alternate pair of angles

âˆ 4 = âˆ 6 =Â 72o

Therefore, âˆ 4 = 72o

âˆ 3 and âˆ 5 are alternate pair of angles

âˆ 3 = âˆ 5 =Â 108o

Therefore, âˆ 3 = 108o

âˆ 2 and âˆ 8 are alternate exterior of angles

âˆ 2 = âˆ 8 =Â 72o

Therefore, âˆ 8 = 72o

âˆ 1 and âˆ 7 are alternate exterior of angles

âˆ 1 = âˆ 7 =Â 108o

Therefore, âˆ 7 = 108o

Hence, âˆ 1 = 108o, âˆ 2 = 72o, âˆ 3 = 108o, âˆ 4 = 72o, âˆ 5 = 108o, âˆ 6 = 72o, âˆ 7 = 108o, âˆ 8 = 72o

12. In Fig. 69, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find âˆ 1, âˆ 2 and âˆ 3.

Â

Solution:

Given l, m and n are parallel lines intersected by transversal p at X, Y and Z

Therefore linear pair,

âˆ 4 + 60o = 180o

âˆ 4 = 180o â€“ 60o

âˆ 4 = 120o

From the figure,

âˆ 4 and âˆ 1 are corresponding angles

âˆ 4 = âˆ 1

Therefore, âˆ 1 = 120o

âˆ 1 and âˆ 2 are corresponding angles

âˆ 2 = âˆ 1

Therefore, âˆ 2 = 120o

âˆ 2 and âˆ 3 are vertically opposite angles

âˆ 2 = âˆ 3

Therefore, âˆ 3 = 1200

13. In Fig. 70, if l âˆ¥ m âˆ¥ n and âˆ 1 = 60o, find âˆ 2

Solution:

Given that l âˆ¥ m âˆ¥ n

From the figure Corresponding angles are

âˆ 1 = âˆ 3

âˆ 1 = 60o

Therefore, âˆ 3 = 60o

âˆ 3 and âˆ 4 are linear pair

âˆ 3 + âˆ 4 = 180o

âˆ 4 = 180o â€“ 60o

âˆ 4 = 120o

âˆ 2 and âˆ 4 are alternate interior angles

âˆ 4 = âˆ 2

Therefore, âˆ 2 = 120o

14. In Fig. 71, if AB âˆ¥ CD and CD âˆ¥ EF, find âˆ ACE

Solution:

Given that, AB âˆ¥ CD and CD âˆ¥ EF

Sum of the interior angles,

âˆ CEF + âˆ ECD = 180o

130o + âˆ ECD = 180o

âˆ ECD = 180o â€“ 130o

âˆ ECD = 50o

We know that alternate angles are equal

âˆ BAC = âˆ ACD

âˆ BAC = âˆ ECD + âˆ ACE

âˆ ACE = 70o â€“ 50o

âˆ ACE = 20o

Therefore, âˆ ACE = 20o

15. In Fig. 72, if l âˆ¥ m, n âˆ¥ p and âˆ 1 = 85o, find âˆ 2.

Solution:

Given that, âˆ 1 = 85o

âˆ 1 and âˆ 3 are corresponding angles

So,Â âˆ 1 = âˆ 3

âˆ 3 = 85o

Sum of the interior angles is 180o

âˆ 3 + âˆ 2 = 180o

âˆ 2 = 180o â€“ 85o

âˆ 2 = 95o

16. In Fig. 73, a transversal n cuts two lines l and m. If âˆ 1 = 70o and âˆ 7 = 80o, is l âˆ¥ m?

Solution:

Given âˆ 1 = 70o and âˆ 7 = 80o

We know that if the alternate exterior angles of the two lines are equal, then the lines are parallel.

Here,Â âˆ 1 and âˆ 7 are alternate exterior angles, but they are not equal

âˆ 1 â‰  âˆ 7

17. In Fig. 74, a transversal n cuts two lines l and m such that âˆ 2 = 65o and âˆ 8 = 65o. Are the lines parallel?

Solution:

From the figure âˆ 2 = âˆ 4 are vertically opposite angles,

âˆ 2 = âˆ 4 = 65o

âˆ 8 = âˆ 6 = 65o

Therefore, âˆ 4 = âˆ 6

Hence, l âˆ¥ m

18. In Fig. 75, Show that AB âˆ¥ EF.

Solution:

We know that,

âˆ ACD = âˆ ACE + âˆ ECD

âˆ ACD = 22o + 35o

âˆ ACD = 57o = âˆ BAC

Thus, lines BA and CD are intersected by the line AC such that, âˆ ACD = âˆ BAC

So, the alternate angles are equal

Therefore, AB âˆ¥ CDÂ â€¦â€¦1

Now,

âˆ ECD + âˆ CEF = 35o + 145o = 180o

This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180o

So, they are supplementary angles

Therefore, EF âˆ¥ CDÂ â€¦â€¦.2

From equation 1 and 2

We conclude that, AB âˆ¥ EF

19. In Fig. 76, AB âˆ¥ CD. Find the values of x, y, z.

Solution:

Given that AB âˆ¥ CD

Linear pair,

âˆ xÂ + 125o = 180o

âˆ xÂ = 180o â€“ 125o

âˆ x = 55o

Corresponding angles

âˆ z = 125o

âˆ x + âˆ z = 180o

âˆ x + 125o = 180o

âˆ x = 180o â€“ 125o

âˆ x = 55o

âˆ x + âˆ y = 180o

âˆ y + 55o = 180o

âˆ y = 180o â€“ 55o

âˆ y = 125o

20. In Fig. 77, find out âˆ PXR, if PQ âˆ¥ RS.

Solution:

Given PQ âˆ¥ RS

We need to find âˆ PXR

âˆ XRS = 50o

âˆ XPQ = 70o

Given, that PQ âˆ¥ RS

âˆ PXR = âˆ XRS + âˆ XPR

âˆ PXR = 50o + 70o

âˆ PXR = 120o

Therefore, âˆ PXR = 120o

21. In Figure, we have

(i) âˆ MLY = 2âˆ LMQ

(ii) âˆ XLM = (2x – 10)o and âˆ LMQ = (x + 30)o, find x.

(iii) âˆ XLM = âˆ PML, find âˆ ALY

(iv) âˆ ALY = (2x – 15)o, âˆ LMQ = (x + 40)o , find x.

Solution:

(i) âˆ MLY and âˆ LMQ are interior angles

âˆ MLY + âˆ LMQ = 180o

2âˆ LMQ + âˆ LMQ = 180o

3âˆ LMQ = 180o

âˆ LMQ = 180o/3

âˆ LMQ = 60o

(ii) âˆ XLM = (2x – 10)o and âˆ LMQ = (x + 30)o, find x.

âˆ XLM = (2x – 10)o and âˆ LMQ = (x + 30)o

âˆ XLM and âˆ LMQ are alternate interior angles

âˆ XLM = âˆ LMQ

(2x – 10)o = (x + 30)o

2x â€“ x = 30o + 10o

x = 40o

Therefore, x = 40Â°

(iii) âˆ XLM = âˆ PML, find âˆ ALY

âˆ XLM = âˆ PML

Sum of interior angles is 180 degrees

âˆ XLM + âˆ PML = 180o

âˆ XLM + âˆ XLM = 180o

2âˆ XLM = 180o

âˆ XLM = 180o/2

âˆ XLM = 90o

âˆ XLM and âˆ ALY are vertically opposite angles

Therefore, âˆ ALY = 90o

(iv) âˆ ALY = (2x – 15)o, âˆ LMQ = (x + 40)o, find x.

âˆ ALY and âˆ LMQ are corresponding angles

âˆ ALY = âˆ LMQ

(2x – 15)o = (x + 40)o

2x â€“ x = 40o + 15o

x = 55o

Therefore, x = 55o

22. In Fig. 79, DE âˆ¥ BC. Find the values of x and y.

Solution:

We know that,

ABC, DAB are alternate interior angles

âˆ ABC = âˆ DAB

So, x = 40o

And ACB, EAC are alternate interior angles

âˆ ACB = âˆ EAC

So, y = 55o

23. In Fig. 80, line AC âˆ¥ line DE and âˆ ABD = 32o, Find out the angles x and y if âˆ E = 122o.

Solution:

Given line AC âˆ¥ line DE and âˆ ABD = 32o

âˆ BDE = âˆ ABD = 32o â€“ Alternate interior angles

âˆ BDE + y = 180oâ€“ linear pair

32o + y = 180o

y = 180o â€“ 32o

y = 148o

âˆ ABE = âˆ E = 122o â€“ Alternate interior angles

âˆ ABD + âˆ DBE = 122o

32o + x = 122o

x = 122o â€“ 32o

x = 90o

24. In Fig. 81, side BC of Î”ABC has been produced to D and CE âˆ¥ BA. If âˆ ABC = 65o, âˆ BAC = 55o, find âˆ ACE, âˆ ECD, âˆ ACD.

Solution:

Given âˆ ABC = 65o, âˆ BAC = 55o

Corresponding angles,

âˆ ABC = âˆ ECD = 65o

Alternate interior angles,

âˆ BAC = âˆ ACE = 55o

Now, âˆ ACD = âˆ ACE + âˆ ECD

âˆ ACD = 55o + 65o

= 120o

25. In Fig. 82, line CA âŠ¥ AB âˆ¥ line CR and line PR âˆ¥ line BD. FindÂ âˆ x, âˆ y, âˆ z.

Solution:

Given that, CA âŠ¥ AB

âˆ CAB = 90o

âˆ AQP = 20o

By, angle of sum property

In Î”ABC

âˆ CAB + âˆ AQP + âˆ APQ = 180o

âˆ APQ = 180o â€“ 90o â€“ 20o

âˆ APQ = 70o

y and âˆ APQ are corresponding angles

y = âˆ APQ = 70o

âˆ APQ and âˆ z are interior angles

âˆ APQ + âˆ z = 180o

âˆ z = 180o â€“ 70o

âˆ z = 110o

26. In Fig. 83, PQ âˆ¥ RS. Find the value of x.

Â

Solution:

Given, linear pair,

âˆ RCD + âˆ RCB = 180o

âˆ RCB = 180o â€“ 130o

= 50o

In Î”ABC,

âˆ BAC + âˆ ABC + âˆ BCA = 180o

By, angle sum property

âˆ BAC = 180o â€“ 55o â€“ 50o

âˆ BAC = 75o

27. In Fig. 84, AB âˆ¥ CD and AE âˆ¥ CF, âˆ FCG = 90o and âˆ BAC = 120o. Find the value of x, y and z.

Solution:

Alternate interior angle

âˆ BAC = âˆ ACG = 120o

âˆ ACF + âˆ FCG = 120o

So, âˆ ACF = 120o â€“ 90o

= 30o

Linear pair,

âˆ DCA + âˆ ACG = 180o

âˆ x = 180o â€“ 120o

= 60o

âˆ BAC + âˆ BAE + âˆ EAC = 360o

âˆ CAE = 360o â€“ 120o â€“ (60o + 30o)

= 150o

28. In Fig. 85, AB âˆ¥ CD and AC âˆ¥ BD. Find the values of x, y, z.

Solution:

(i)Â  Since, AC âˆ¥ BD and CD âˆ¥ AB, ABCD is a parallelogram

âˆ CAB + âˆ ACD = 180o

âˆ ACD = 180o â€“ 65o

= 115o

Opposite angles of parallelogram,

âˆ CAB = âˆ CDB = 65o

âˆ ACD = âˆ DBA = 115o

(ii)Â  Here,

AC âˆ¥ BD and CD âˆ¥ AB

Alternate interior angles,

âˆ CAD = x = 40o

âˆ DAB = y = 35o

29. In Fig. 86, state which lines are parallel and why?

Solution:

Let, F be the point of intersection of the line CD and the line passing through point E.

Here, âˆ ACD and âˆ CDE are alternate and equal angles.

So, âˆ ACD = âˆ CDEÂ = 100o

Therefore, AC âˆ¥ EF

30. In Fig. 87, the corresponding arms of âˆ ABC and âˆ DEF are parallel. If âˆ ABC = 75o, find âˆ DEF.

Solution:

Let, G be the point of intersection of the lines BC and DE

Since, AB âˆ¥ DE and BC âˆ¥ EF

The corresponding angles are,

âˆ ABC = âˆ DGC = âˆ DEF = 75o