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

âˆ BAC and âˆ CAD

In fig (ii), the adjacent angles are

âˆ BAD and âˆ DAC

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) 35 ^{o}**

**(ii) 72 ^{o}**

**(iii) 45 ^{o}**

**(iv) 85 ^{o}**

**Solution:**

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

Complementary angle for given angle is

90^{o} â€“ 35^{o} = 55^{o}

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

Complementary angle for given angle is

90^{0} â€“ 72^{o} = 18^{o}

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

Complementary angle for given angle is

90^{o }â€“ 45^{o} = 45^{o}

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

Complementary angle for given angle is

90^{o} â€“ 85^{o} = 5^{o}

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

**(i) 70 ^{o}**

**(ii) 120 ^{o}**

**(iii) 135 ^{o}**

**(iv) 90 ^{o}**

**Solution:**

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

Therefore supplementary angle for the given angle is

180^{o} â€“ 70^{o} = 110^{o}

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

Therefore supplementary angle for the given angle is

180^{o} â€“ 120^{o} = 60^{o}

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

Therefore supplementary angle for the given angle is

180^{o} â€“ 135^{o} = 45^{o}

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

Therefore supplementary angle for the given angle is

180^{o} â€“ 90^{o} = 90^{o}

Â

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

**(i) 25 ^{o}, 65^{o}**

**(ii) 120 ^{o}, 60^{o}**

**(iii) 63 ^{o}, 27^{o}**

**(iv) 100 ^{o}, 80^{o}**

**Solution:**

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

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

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

(iv) 100^{o} + 80^{o} = 180^{o} 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 90^{o} so their sum will be greater than 180^{o}

(ii) Yes, two right angles can be supplementary

Because, 90^{o} + 90^{o} = 180^{o}

(iii) No, two acute angle cannot be supplementary

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

**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 180^{o}

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 180^{o}

âˆ p = âˆ q

So,

âˆ p + âˆ q = 180^{o}

âˆ p + âˆ p = 180^{o}

2âˆ p = 180^{o}

âˆ p = 180^{o}/2

âˆ p = 90^{o}

Therefore, âˆ p = âˆ q = 90^{o}

**12. If the complement of an angle is 28 ^{o}, then find the supplement of the angle.**

**Solution:**

Given complement of an angle is 28^{o}

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

Therefore, âˆ x + 28^{o} = 90^{o}

âˆ x = 90^{o} â€“ 28^{o}

= 62^{o}

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

= 118^{o}

**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 = 70 ^{o}, find the magnitude of âˆ 2, âˆ 3 and âˆ 4. **

** **

** **

**Solution:**

Given, âˆ 1 = 70^{o}

âˆ 3 = 2(âˆ 1)

= 2(70^{o})

âˆ 3 = 140^{o}

âˆ 3 = âˆ 4

As, OE is the angle bisector,

âˆ DOB = 2(âˆ 1)

= 2(70^{o})

= 140^{o}

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

140^{o} + 140^{o} + 2(âˆ COB) = 360^{o}

Since, âˆ COB = âˆ AOD

2(âˆ COB) = 360^{o} â€“ 280^{o}

2(âˆ COB) = 80^{o}

âˆ COB = 80^{o}/2

âˆ COB = 40^{o}

Therefore, âˆ COB = âˆ AOD = 40^{o}

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

**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 90^{o}

We know that linear pair angle is 180^{o}

Therefore, the other angle is

180^{o} â€“ 90^{o} = 90^{o}

**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 180^{o}.

**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 180^{o}.

**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 180^{o}.

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

**Solution:**

Let x be the required angle

So, x + 65^{o} = 180^{o}

x = 180^{o} â€“ 65^{o}

x = 115^{o}

The two angles are said to be complementary angles if the sum of those angles is 90^{o} here it is more than 90^{o} 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 = 180^{o}

^{o}]

60^{o} + x^{o} = 180^{o}

x^{o} = 180^{o} â€“ 60^{o}

x^{o} = 120^{o}

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

^{o}]

3x^{o} + 2x^{o} = 180^{o}

5x^{o} = 180^{o}

x^{o} = 180^{o}/5

x^{o} = 36^{o}

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

^{o}]

Since, 35^{o} + x^{o} + 60^{o} = 180^{o}

x^{o} = 180^{o} â€“ 35^{o} â€“ 60^{o}

x^{o} = 180^{o} â€“ 95^{o}

x^{o} = 85^{o}

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

83^{o} + 92^{o} + 47^{o} + 75^{o} + x^{o} = 360^{o}

x^{o} + 297^{o} = 360^{o}

x^{o} = 360^{o} â€“ 297^{o}

x^{o} = 63^{o}

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

3x^{o} + 2x^{o} + x^{o} + 2x^{o} = 360^{o}

8x^{o} = 360^{o}

x^{o} = 360^{o}/8

x^{o} = 45^{o}

(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 180^{o}

Therefore 3x^{o} = 105^{o}

x^{o} = 105^{o}/3

x^{o} = 35^{o}

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

** **

**Solution:**

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

Therefore, âˆ 3 = 65^{o}

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 180^{o}]

Therefore, âˆ 2 = 180^{o} â€“ 65^{o}

= 115^{o}

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

Therefore, âˆ 2 = âˆ 4 = 115^{o}

And âˆ 3 = 65^{o}

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

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

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

**Solution:**

(i) âˆ AOC + âˆ BOC = 180^{o} [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 180^{o}]

2y + 5^{0} + 3x = 180^{o}

3x + 2y = 175^{o}

Given If x = 25^{o}, then

3(25^{o}) + 2y = 175^{o}

75^{o} + 2y = 175^{o}

2y = 175^{o} â€“ 75^{o}

2y = 100^{o}

y = 100^{o}/2

y = 50^{o}

(ii) âˆ AOC + âˆ BOC = 180^{o} [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 180^{o}]

2y + 5 + 3x = 180^{o}

3x + 2y = 175^{o}

Given If y = 35^{o}, then

3x + 2(35^{o}) = 175^{o}

3x + 70^{o} = 175^{o}

3x = 175^{0} â€“ 70^{o}

3x = 105^{o}

x = 105^{o}/3

x = 35^{o}

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

** **

**Solution:**

Pairs of adjacent angles are:

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 180^{o}]

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} + x^{o} + (x + 20)^{o} = 180^{o}[linear pair]

On rearranging we get

3x^{o} + 30^{o} = 180^{o}

3x^{o} = 180^{o} â€“ 30^{o}

3x^{o} = 150^{o}

x^{o} = 150^{o}/3

x^{o} = 50^{o}

Also given that

âˆ BOC = (x + 20)^{o}

= (50 + 20)^{o}

= 70^{o}

âˆ COD = 50^{o}

âˆ AOD = (x + 10)^{o}

= (50 + 10)^{o}

= 60^{o}

**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 = 180^{o }[linear pair]

3x^{o} + 3x^{o} = 180^{o}

6x^{o} = 180^{o}

x^{o} = 180^{o}/6

x^{o} = 30^{o}

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

** **

**Solution: **

From the figure we can write as

âˆ AOB + âˆ BOC = 180^{o} [linear pair]

Linear pair

2x + 70^{o} = 180^{o}

2x = 180^{o} â€“ 70^{o}

2x = 110^{o}

x = 110^{o}/2

x = 55^{o}

**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 = 180^{o}

60^{o} + 4x + 40^{o} = 180^{o}

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

4x = 180^{o} â€“ 100^{o}

4x = 80^{o}

x = 80^{o}/4

x = 20^{o}

**30. In Fig. 30, lines l _{1 }and l_{2} intersect at O, forming angles as shown in the figure. If x = 45^{o}, find the values of y, z and u.**

** **

**Solution:**

Given that, âˆ x = 45^{o}

From the figure we can write as

âˆ x = âˆ z = 45^{o}

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 = 360^{o}

45^{o} + 45^{o} + âˆ y + âˆ u = 360^{o}

90^{o} + âˆ y + âˆ u = 360^{o}

âˆ y + âˆ u = 360^{o} â€“ 90^{o}

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

2âˆ y = 270^{o}

âˆ y = 135^{o}

Therefore, âˆ y = âˆ u = 135^{o}

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

**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 + 50^{o} + 90^{o} = 360^{o}

Linear pair, âˆ x + 50^{o} + 90^{o} = 180^{o}

âˆ x + 140^{o} = 180^{o}

On rearranging we get

âˆ x = 180^{o} â€“ 140^{o}

âˆ x = 40^{o}

From the figure we can write as

âˆ x = âˆ u = 40^{o} are vertically opposite angles

âˆ z = 90^{o} is a vertically opposite angle

âˆ y = 50^{o} is a vertically opposite angle

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

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

** **

**Solution:**

âˆ y = 25^{o} vertically opposite angle

From the figure we can write as

âˆ x = âˆ z are vertically opposite angles

âˆ x + âˆ y + âˆ z + 25^{o} = 360^{o}

âˆ x + âˆ z + 25^{o} + 25^{o} = 360^{o}

On rearranging we get,

âˆ x + âˆ z + 50^{o} = 360^{o}

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

2âˆ x = 310^{o}

âˆ x = 155^{o}

And, âˆ x = âˆ z = 155^{o}

Therefore, âˆ x = 155^{o}, âˆ y = 25^{o} and âˆ z = 155^{o}

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 = 60 ^{o}, 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 = 60^{o} [given]

Vertically opposite angles are

âˆ LMD = âˆ CMQ = 60^{o }[given]

Vertically opposite angles are

âˆ ALM = âˆ PLB = 60^{o}

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

On rearranging we get

âˆ QMD = 180^{o} â€“ 60^{o}

= 120^{o}

Corresponding angles are

âˆ QMD = âˆ MLB = 120^{o}

Vertically opposite angles

âˆ QMD = âˆ CML = 120^{o}

Vertically opposite angles

âˆ MLB = âˆ ALP = 120^{o}

**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 = 35 ^{o} find âˆ ALM and âˆ PLA.**

**Solution:**

Given that, âˆ LMD = 35^{o}

From the figure we can write

âˆ LMD and âˆ LMC is a linear pair

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

On rearranging, we get

âˆ LMC = 180^{o} â€“ 35^{o}

= 145^{o}

So, âˆ LMC = âˆ PLA = 145^{o}

And, âˆ LMC = âˆ MLB = 145^{o}

âˆ MLB and âˆ ALM is a linear pair

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

âˆ ALM = 180^{o} â€“ 145^{o}

âˆ ALM = 35^{0}

Therefore, âˆ ALM = 35^{o}, âˆ PLA = 145^{o}.

**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 = 40^{o}

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

âˆ 1 + âˆ 2 = 180^{o}

âˆ 2 = 180^{o} â€“ 40^{o}

âˆ 2 = 140^{o}

Again from the figure we can say that

âˆ 2 and âˆ 6Â is a corresponding angle pair

So, âˆ 6 = 140^{o}

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

âˆ 6 + âˆ 5 = 180^{o}

âˆ 5 = 180^{o} â€“ 140^{o}

âˆ 5 = 40^{o}

From the figure we can write as

âˆ 3 and âˆ 5Â are alternate interior angles

So, âˆ 5 = âˆ 3 = 40^{o}

âˆ 3 and âˆ 4 is a linear pair

âˆ 3 + âˆ 4 = 180^{o}

âˆ 4 = 180^{o} â€“ 40^{o}

âˆ 4 = 140^{o}

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

So, âˆ 4 = âˆ 6 = 140^{o}

âˆ 3 and âˆ 7 are a pair of corresponding angles

So, âˆ 3 = âˆ 7 = 40^{o}

Therefore, âˆ 7 = 40^{o}

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

So, âˆ 4 = âˆ 8 = 140^{o}

Therefore, âˆ 8 = 140^{o}

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

**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 = 75^{o}

âˆ 1 = âˆ 3 are vertically opposite angles

We know that, from the figure

âˆ 1 + âˆ 2 = 180^{o} is a linear pair

âˆ 2 = 180^{o} â€“ 75^{o}

âˆ 2 = 105^{o}

Here, âˆ 1 = âˆ 5 = 75^{o} are corresponding angles

âˆ 5 = âˆ 7 = 75^{o} are vertically opposite angles.

âˆ 2 = âˆ 6 = 105^{o} are corresponding angles

âˆ 6 = âˆ 8 = 105^{o} are vertically opposite angles

âˆ 2 = âˆ 4 = 105^{o} are vertically opposite angles

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

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

** **

**Solution:**

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

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

âˆ QMC = 180^{o} â€“ âˆ QMD

âˆ QMC = 180^{o} â€“ 100Â°

âˆ QMC = 80^{o}

Corresponding angles are

âˆ DMQ = âˆ BLM = 100^{o}

âˆ CMQ = âˆ ALM = 80^{o}

Vertically Opposite angles are

âˆ DMQ = âˆ CML = 100^{o}

âˆ BLM = âˆ PLA = 100^{o}

âˆ CMQ = âˆ DML = 80^{o}

âˆ ALM = âˆ PLB = 80^{o}

**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 80^{o}

âˆ z and 80^{o} are vertically opposite angles

Therefore âˆ z = 80^{o}

âˆ z and âˆ t are corresponding angles

âˆ z = âˆ t

Therefore, âˆ t = 80^{o}

âˆ z and âˆ y are corresponding angles

âˆ z = âˆ y

Therefore, âˆ y = 80^{o}

âˆ x and âˆ y are corresponding angles

âˆ y = âˆ x

Therefore, âˆ x = 80^{o}

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

**Solution:**

Given that, âˆ 1 = 120^{o} and âˆ 2 = 100^{o}

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

âˆ 1 + âˆ 5 = 180^{o}

âˆ 5 = 180^{o} â€“ 120^{o}

âˆ 5 = 60^{o}

Therefore, âˆ 5 = 60^{o}

âˆ 2 and âˆ 6 are corresponding angles

âˆ 2 = âˆ 6 =Â 100^{o}

Therefore,Â âˆ 6 =Â 100^{o}

âˆ 6 and âˆ 3 a linear pair

âˆ 6 + âˆ 3 = 180^{o}

âˆ 3 = 180^{o} â€“ 100^{o}

âˆ 3 = 80^{o}

Therefore, âˆ 3 = 80^{o}

By, angles of sum property

âˆ 3 + âˆ 5 + âˆ 4Â = 180^{o}

âˆ 4 = 180^{o} â€“ 80^{o} â€“ 60^{o}

âˆ 4 = 40^{o}

Therefore, âˆ 4 = 40^{o}

**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 = 110^{o}

Corresponding angles, âˆ a = âˆ b

Therefore, âˆ b = 110^{o}

Vertically opposite angle,

âˆ d = 85^{o}

Corresponding angles, âˆ d = âˆ c

Therefore, âˆ c = 85^{o}

Hence, âˆ a = 110^{o}, âˆ b = 110^{o}, âˆ c = 85^{o}, âˆ d = 85^{o}

**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 = 180^{o}

5x = 180^{o}

x = 180^{o}/5

x = 36^{o}

Therefore, âˆ 1 = 3x = 3(36) = 108^{o}

âˆ 2 = 2x = 2(36) = 72^{o}

âˆ 1 and âˆ 5 are corresponding angles

Therefore âˆ 1 = âˆ 5

Hence, âˆ 5 = 108^{o}

âˆ 2 and âˆ 6 are corresponding angles

So âˆ 2 = âˆ 6

Therefore, âˆ 6 = 72^{o}

âˆ 4 and âˆ 6 are alternate pair of angles

âˆ 4 = âˆ 6 =Â 72^{o}

Therefore, âˆ 4 = 72^{o}

âˆ 3 and âˆ 5 are alternate pair of angles

âˆ 3 = âˆ 5 =Â 108^{o}

Therefore, âˆ 3 = 108^{o}

âˆ 2 and âˆ 8 are alternate exterior of angles

âˆ 2 = âˆ 8 =Â 72^{o}

Therefore, âˆ 8 = 72^{o}

âˆ 1 and âˆ 7 are alternate exterior of angles

âˆ 1 = âˆ 7 =Â 108^{o}

Therefore, âˆ 7 = 108^{o}

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

Â

**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 + 60^{o} = 180^{o}

âˆ 4 = 180^{o} â€“ 60^{o}

âˆ 4 = 120^{o}

From the figure,

âˆ 4 and âˆ 1 are corresponding angles

âˆ 4 = âˆ 1

Therefore, âˆ 1 = 120^{o}

âˆ 1 and âˆ 2 are corresponding angles

âˆ 2 = âˆ 1

Therefore, âˆ 2 = 120^{o}

âˆ 2 and âˆ 3 are vertically opposite angles

âˆ 2 = âˆ 3

Therefore, âˆ 3 = 120^{0}

Â

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

**Solution:**

Given that l âˆ¥ m âˆ¥ n

From the figure Corresponding angles are

âˆ 1 = âˆ 3

âˆ 1 = 60^{o}

Therefore, âˆ 3 = 60^{o}

âˆ 3 and âˆ 4 are linear pair

âˆ 3 + âˆ 4 = 180^{o}

âˆ 4 = 180^{o} â€“ 60^{o}

âˆ 4 = 120^{o}

âˆ 2 and âˆ 4 are alternate interior angles

âˆ 4 = âˆ 2

Therefore, âˆ 2 = 120^{o}

**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 = 180^{o}

130^{o} + âˆ ECD = 180^{o}

âˆ ECD = 180^{o} â€“ 130^{o}

âˆ ECD = 50^{o}

We know that alternate angles are equal

âˆ BAC = âˆ ACD

âˆ BAC = âˆ ECD + âˆ ACE

âˆ ACE = 70^{o} â€“ 50^{o}

âˆ ACE = 20^{o}

Therefore, âˆ ACE = 20^{o}

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

**Solution:**

Given that, âˆ 1 = 85^{o}

âˆ 1 and âˆ 3 are corresponding angles

So,Â âˆ 1 = âˆ 3

âˆ 3 = 85^{o}

Sum of the interior angles is 180^{o}

âˆ 3 + âˆ 2 = 180^{o}

âˆ 2 = 180^{o} â€“ 85^{o}

âˆ 2 = 95^{o}

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

**Solution:**

Given âˆ 1 = 70^{o} and âˆ 7 = 80^{o}

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 = 65 ^{o} and âˆ 8 = 65^{o}. Are the lines parallel?**

**Solution:**

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

âˆ 2 = âˆ 4 = 65^{o}

âˆ 8 = âˆ 6 = 65^{o}

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 = 22^{o} + 35^{o}

âˆ ACD = 57^{o} = 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 = 35^{o} + 145^{o} = 180^{o}

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

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 + 125^{o} = 180^{o}

âˆ xÂ = 180^{o} â€“ 125^{o}

âˆ x = 55^{o}

Corresponding angles

âˆ z = 125^{o}

Adjacent interior angles

âˆ x + âˆ z = 180^{o}

âˆ x + 125^{o} = 180^{o}

âˆ x = 180^{o} â€“ 125^{o}

âˆ x = 55^{o}

Adjacent interior angles

âˆ x + âˆ y = 180^{o}

âˆ y + 55^{o} = 180^{o}

âˆ y = 180^{o} â€“ 55^{o}

âˆ y = 125^{o}

**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 = 50^{o}

âˆ XPQ = 70^{o}

Given, that PQ âˆ¥ RS

âˆ PXR = âˆ XRS + âˆ XPR

âˆ PXR = 50^{o} + 70^{o}

âˆ PXR = 120^{o}

Therefore, âˆ PXR = 120^{o}

**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 = 180^{o}

2âˆ LMQ + âˆ LMQ = 180^{o}

3âˆ LMQ = 180^{o}

âˆ LMQ = 180^{o}/3

âˆ LMQ = 60^{o}

(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 = 30^{o} + 10^{o}

x = 40^{o}

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 = 180^{o}

âˆ XLM + âˆ XLM = 180^{o}

2âˆ XLM = 180^{o}

âˆ XLM = 180^{o}/2

âˆ XLM = 90^{o}

âˆ XLM and âˆ ALY are vertically opposite angles

Therefore, âˆ ALY = 90^{o}

(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 = 40^{o} + 15^{o}

x = 55^{o}Â

Therefore, x = 55^{o}

**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 = 40^{o}

And ACB, EAC are alternate interior angles

âˆ ACB = âˆ EAC

So, y = 55^{o}

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

**Solution:**

Given line AC âˆ¥ line DE and âˆ ABD = 32^{o}

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

âˆ BDE + y = 180^{o}â€“ linear pair

32^{o }+ y = 180^{o}

y = 180^{o} â€“ 32^{o}

y = 148^{o}

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

âˆ ABD + âˆ DBE = 122^{o}

32^{o} + x = 122^{o}

x = 122^{o} â€“ 32^{o}

x = 90^{o}

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

**Solution:**

Given âˆ ABC = 65^{o}, âˆ BAC = 55^{o}

Corresponding angles,

âˆ ABC = âˆ ECD = 65^{o}

Alternate interior angles,

âˆ BAC = âˆ ACE = 55^{o}

Now, âˆ ACD = âˆ ACE + âˆ ECD

âˆ ACD = 55^{o} + 65^{o}

= 120^{o}

**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 = 90^{o}

âˆ AQP = 20^{o}

By, angle of sum property

In Î”ABC

âˆ CAB + âˆ AQP + âˆ APQ = 180^{o}

âˆ APQ = 180^{o} â€“ 90^{o} â€“ 20^{o}

âˆ APQ = 70^{o}

y and âˆ APQ are corresponding angles

y = âˆ APQ = 70^{o}

âˆ APQ and âˆ z are interior angles

âˆ APQ + âˆ z = 180^{o}

âˆ z = 180^{o} â€“ 70^{o}

âˆ z = 110^{o}

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

**Â **

**Solution:**

Given, linear pair,

âˆ RCD + âˆ RCB = 180^{o}

âˆ RCB = 180^{o} â€“ 130^{o}

= 50^{o}

In Î”ABC,

âˆ BAC + âˆ ABC + âˆ BCA = 180^{o}

By, angle sum property

âˆ BAC = 180^{o} â€“ 55^{o} â€“ 50^{o}

âˆ BAC = 75^{o}

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

**Solution:**

Alternate interior angle

âˆ BAC = âˆ ACG = 120^{o}

âˆ ACF + âˆ FCG = 120^{o}

So, âˆ ACF = 120^{o} â€“ 90^{o}

= 30^{o}

Linear pair,

âˆ DCA + âˆ ACG = 180^{o}

âˆ x = 180^{o} â€“ 120^{o}

= 60^{o}

âˆ BAC + âˆ BAE + âˆ EAC = 360^{o}

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

= 150^{o}

**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

Adjacent angles of parallelogram,

âˆ CAB + âˆ ACD = 180^{o}

âˆ ACD = 180^{o} â€“ 65^{o}

= 115^{o}

Opposite angles of parallelogram,

âˆ CAB = âˆ CDB = 65^{o}

âˆ ACD = âˆ DBA = 115^{o}

(ii)Â Here,

AC âˆ¥ BD and CD âˆ¥ AB

Alternate interior angles,

âˆ CAD = x = 40^{o}

âˆ DAB = y = 35^{o}

**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 = 100^{o}

Therefore, AC âˆ¥ EF

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