# RD Sharma Solutions Class 7 Chapter 14 Line and Angles

Read RD Sharma Solutions Class 7 Chapter 14 Line and Angles below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations

Exercise 14.1

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

Solution 1:

Adjacent angles are those that have a common vertex and a common arm.

As a result, the adjacent angles in the diagram are:

∠DOC and ∠BOC

∠COB and ∠BOA

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

Solution 2:

Adjacent angles are those that have a common vertex and a common arm.

In fig (i), the adjacent angles are

∠EBA and ∠ABC

∠ACB and ∠BCF

In fig (ii), the adjacent angles are

∠BDA and ∠CDA

Question :3. In fig. 15, write down

(i) Each linear pair

(ii) Each pair of vertically opposite angles.

Solution 3:
(i) If the non-common arms of two adjacent angles are two opposite rays, they are said to form a linear pair of angles.
∠1 and ∠3 ∠5 and ∠6
∠1 and ∠2 ∠5 and ∠7
∠4 and ∠3 ∠6 and ∠8
∠4 and ∠2 ∠7 and ∠8

(ii) The two angles created by two intersecting lines that have no common arms are called vertically opposite angles.
∠1 and ∠4 ∠5 and ∠8
∠2 and ∠3 ∠6 and ∠7

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

Solution 4: No, because they don’t have any common vertex.

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

(i) 35o

(ii) 72o

(iii) 45o

(iv) 85o

Solution 5:

The sum of two angles is 90 o then it is a complementary angle.

(i) Thus, complementary angle for given angle is

90o – 35o = 55o

(ii) Thus, complementary angle for given angle is

900 – 72o = 18o

(iii) Thus, complementary angle for given angle is

90– 45o = 45o

(iv) Thus, complementary angle for given angle is

90o – 85o = 5o

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

(i) 70o

(ii) 120o

(iii) 135o

(iv) 90o

Solution 6:

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

(i) Thus, supplementary angle for the given angle is

180o – 70o = 110o

(ii) Thus, supplementary angle for the given angle is

180o – 120o = 60o

(iii) Thus, supplementary angle for the given angle is

180o – 135o = 45o

(iv) Thus, supplementary angle for the given angle is

180o – 90o = 90o

Question :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 7:

(i) 25o + 65o = 90o

So, this is a pair of complementary angles.

(ii) 120o + 60o = 180o

So, this is a pair of supplementary angles.

(iii) 63o + 27o = 90o

So, this is a pair of complementary angles.

(iv) 100o + 80o = 180o

So, this is a pair of supplementary angles.

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

(i) Obtuse?

(ii) Right?

(iii) Acute?

Solution 8:

(i) No, two obtuse angles can't be supplementary.

Since the number of two angles is greater than 90 o, their total is greater than 180o.

(ii) Yes, two right angles can be supplementary

Because, 90o + 90o = 180o

(iii No, two obtuse angles can't be supplementary.

Since the number of two angles is greater than 90 o, their sum will also be less than 90o

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

Solution 9:

If the sum of the two angles is 180o, they are said to be supplementary angles

The supplementary angles are

∠AOC and ∠COB

∠BOD and ∠DOA

∠BOC and ∠DOB

∠AOC and ∠DOA

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

(i) Name two linear pairs.

(ii) Name two pairs of supplementary angles.

Solution 10:

(i) If the non-common arms of two adjacent angles are two opposite rays, they are said to form a linear pair of angles.

Thus, linear pairs are:

∠ABD and ∠DBC

∠ABE and ∠EBC

(ii) Every linear pair form supplementary angle.

Thus, Supplementary angles are:

∠ABD and ∠DBC

∠ABE and ∠EBC

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

Solution  11:

Let p and q be the two supplementary angles that are equal to 180 o

∠p = ∠q

So,

∠p + ∠q = 180o

∠p + ∠p = 180o

2∠p = 180o

∠p =

∠p = 90o

Thus, ∠p = ∠q = 90o

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

Solution  12:

According to the question complement of an angle is 28o

Let x be the complement of the angle 28o

So, ∠x + 28o = 90o

∠x = 90o – 28o = 62o

So, the supplement of the angle = 180o – 62o = 118o

Question :13. In Fig. 19, name each linear pair and each pair of vertically opposite angles:

Solution 13:

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

Thus, linear pairs are listed below:

∠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.

Thus, supplement of the angle is listed below:

∠1 and ∠3                           ∠4 and ∠2                           ∠5 and ∠7

∠6 and ∠8                           ∠9 and ∠11                         ∠10 and ∠12

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

Solution 14:

Given in the question, ∠1 = 70o

∠3 = 2(∠1)

= 2(70o)

∠3 = 140o

∠3 = ∠4

As, OE is the angle bisector,

∠DOB = 2(∠1)

∠DOB = 2(70o)

∠DOB = 140o

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

140o + 140o + 2(∠COB) = 360o

Since, ∠COB = ∠AOD

2(∠COB) = 360o – 280o

2(∠COB) = 80o

∠COB = 80o/2

∠COB = 40o

Thus, ∠COB = ∠AOD = 40o

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

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

Solution 15:

According to the question one of the angle of a linear pair is the right angle that is 90o

Linear pair angle is 180o

180o – 90o = 90o

So, the other angle is 90o.

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

Solution 16:

According to the question If one of the angles in a linear pair is obtuse, then the other must be acute, since only then can the sum be 180o.

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

Solution 17:

According to the question one of the angles in a linear pair is acute, then the other must be obtuse; only then will their sum be 180o.

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

Solution 18:

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

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

Solution 19:

Let x be the required angle

So, x + 65o = 180o

x = 180o – 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.

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

Solution 20:

(i)   ∠BOA + ∠BOC = 180o

60o + xo = 180o

xo = 180o – 60o

xo = 120o

(ii) ∠POQ + ∠QOR = 180o

3xo + 2xo = 180o

5xo = 180o

xo = 180o/5

xo = 36o

(iii) ∠LOP + ∠PON + ∠NOM = 180o

Since, 35o + xo + 60o = 180o

xo = 180o – 35o – 60o

xo = 180o – 95o

xo = 85o

(iv) ∠DOC + ∠DOE + ∠EOA + ∠AOB+ ∠BOC = 360o

83o + 92o + 47o + 75o + xo = 360o

xo + 297o = 360o

xo = 360o – 297o

xo = 63o

(v) ∠ROS + ∠ROQ + ∠QOP + ∠POS = 360o

3xo + 2xo + xo + 2xo = 360o

8xo = 360o

xo = 360o/8

xo = 45o

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

Therefore 3xo = 105o

xo = 105o/3

xo = 35o

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

Solution 21:

According to figure, the vertically opposite angles are ∠2, ∠1 = ∠3

Thus, ∠3 = 65o

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

The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180o

Thus, ∠2 = 180o – 65o = 115o

∠2 = ∠4 are the vertically opposite angles [in the figure]

Therefore, ∠2 = ∠4 = 115o

And ∠1 = 65o

Question :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 22:

(i) ∠AOC + ∠BOC = 180o

The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180o

2y + 50 + 3x = 180o

3x + 2y = 180o - 50

3x + 2y = 175o

Given in the question, If x = 25o, then

3(25o) + 2y = 175o

75o + 2y = 175o

2y = 175o – 75o

2y = 100o

y =  = 50o

(ii) ∠AOC + ∠BOC = 180o

The two adjacent angles are said to form a linear pair of angles if their non–common arms are two opposite rays and sum of the angle is 180o

2y + 5 + 3x = 180o

3x + 2y = 180o - 50

3x + 2y = 175o

Given in the question, If y = 35o, then

3x + 2(35o) = 175o

3x + 70o = 175o

3x = 1750 – 70o

3x = 105o

x =

x = 35o

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

Solution 23:

∠DOA and ∠DOC                             ∠BOC and ∠COD

∠AOD and ∠BOD                             ∠AOC and ∠BOC

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

∠AOD and ∠BOD                             ∠AOC and ∠BOC

Question :24. In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.

Solution 24:

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

By solving the equation

3xo + 30o = 180o

3xo = 180o – 30o

3xo = 150o

xo =

xo = 50o

Also, ∠BOC = (x + 20)o [x=50]

∠BOC = (50 + 20)o = 70o

∠COD = 50o

∠AOD = (x + 10)o

∠AOD = (50 + 10)o = 60o

Question :25. How many pairs of adjacent angles are formed when two lines intersect in a point?

Solution 25:

If two lines intersect at a same 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 Fig. 26?

Solution 26:

There are 10 adjacent pairs formed by the figure 26:

∠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 Fig. 27, determine the value of x.

Solution 27:

According to the figure we can written

∠COB + ∠AOB = 180[linear pair]

3xo + 3xo = 180o

6xo = 180o

xo =   = 30o

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

Solution 28:

According to the figure we can written,

∠AOB + ∠BOC = 180o [linear pair]

2x + 70o = 180o

2x = 180o – 70o

2x = 110o

x =     = 55o

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

Solution 29:

According to the figure we can written

∠QOP + ∠QOR + ∠ROS = 180o                               [straight line angle]

60o + 4x + 40o = 180o

By solving the equation

100o + 4x = 180o

4x = 180o – 100o

4x = 80o

x =     = 20o

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

Solution 30:

In the question given that, ∠x = 45o

According to the figure we can written

∠x = ∠z = 45o

Also, we have

∠y = ∠u

Since the property of linear pair,

=∠x + ∠y + ∠z + ∠u = 360o

=45o + 45o + ∠y + ∠u = 360o

=90o + ∠y + ∠u = 360o

=∠y + ∠u = 360o – 90o

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

=2∠y = 270o

=∠y =

=∠y = 135o

∠y = ∠u = 135o

Therefore, ∠x = 45o, ∠y = 135o, ∠z = 45o and ∠u = 135o

Question :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 31:

In the question given that, ∠x + ∠y + ∠z+ ∠u + 50o + 90o = 360o

∠x + 50o + 90o = 180o                                    [linear pair]

∠x + 140o = 180o

By solving the equation

∠x = 180o – 140o = 40o

According the figure we written as

∠x = ∠u = 40o                                  [vertically opposite angles]

∠z = 90o                                                 [vertically opposite angle]

∠y = 50o                                                 [vertically opposite angle]

Thus, ∠x = 40o, ∠y = 50o, ∠z = 90o and ∠u = 40o

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

Solution 32:

∠y = 25o [vertically opposite angle]

According to the figure we written as

∠x = ∠z                                  [vertically opposite angles]

∠x + ∠y + ∠z + 25o = 360o

∠x + ∠z + 25o + 25o = 360o

By solving the equation

∠x + ∠z + 50o = 360o

∠x + ∠z = 360o – 50o [∠x = ∠z]

2∠x = 310o

∠x = 155o

And, ∠x = ∠z = 155o

Thus, ∠x = 155o, ∠y = 25o and ∠z = 155o

Exercise 14.2

Question :1. In Fig. 58, 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 ∠d and ∠g and angles corresponding to ∠f and ∠h in Fig. 58 (ii)

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

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

Solution 1:

(i) A pair of corresponding angles is a pair of angles in which one arm of each angle is on the same side of the transversal and their other arms are pointed in the same direction.

In Figure (i) Corresponding angles are

∠EGB and ∠GHD                               ∠HGB and ∠FHD

∠EGA and ∠GHC                               ∠AGH and ∠CHF

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

The alternate angles are:

∠EGB and ∠CHF                 ∠HGB and ∠CHG

∠EGA and ∠FHD                                ∠AGH and ∠GHD

(ii) In Figure (ii)

The alternate angle ∠d = ∠e.

The alternate angle to ∠g = ∠b.

The corresponding angle ∠f = ∠c.

The corresponding angle ∠h = ∠a.

(iii) In Figure (iii)

Angle alternate ∠PQR = ∠QRA.

Angle corresponding ∠RQF = ∠ARB.

Angle alternate ∠POE = ∠ARB.

(iv) In Figure (ii)

Pair of interior angles are

∠a = ∠e.                                ∠d = ∠f.

Pair of exterior angles are

∠b = ∠h.                               ∠c = ∠g.

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

Solution 2:

A pair of corresponding angles is a pair of angles in which one arm of each angle is on the same side of the transversal and their other arms are pointed in the same direction.

Thus, corresponding angles are

∠ALM = ∠CMQ = 60o [given in question]

Vertically opposite angles ∠LMD = ∠CMQ = 60[given in question]

Vertically opposite angles ∠ALM = ∠PLB = 60o

Now, ∠CMQ + ∠QMD = 180o are the linear pair

By solving the equation

∠QMD = 180o – 60o = 120o

Corresponding angles are

∠QMD = ∠MLB = 120o

Vertically opposite angles

∠QMD = ∠CML = 120o

∠MLB = ∠ALP = 120o

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

Solution 3:

In the question given that, ∠LMD = 35o

According to the figure we written

∠LMD and ∠LMC [linear pair]

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

By solving the equation

∠LMC = 180o – 35o = 145o

So, ∠LMC = ∠PLA = 145o

And, ∠LMC = ∠MLB = 145o

∠MLB and ∠ALM [linear pair]

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

∠ALM = 180o – 145o = 350

Thus, ∠ALM = 35o, ∠PLA = 145o.

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

Solution 4:

According to the question, l ∥ m

According to the figure the angle alternate to ∠13 = ∠7

According to the figure the angle corresponding to ∠15 = ∠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 ∠15 = ∠5

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

Solution 5:

In the question given that, ∠1 = 40o

∠1 and ∠2              [linear pair]

∠1 + ∠2 = 180o

∠2 = 180o – 40o

∠2 = 140o

Again, from the figure:

∠2 and ∠6            [corresponding angle pair]

So, ∠6 = 140o

∠6 and ∠5            [linear pair]

∠6 + ∠5 = 180o

∠5 = 180o – 140o = 40o

According to the figure we written as

∠3 and ∠5                            [alternate interior angles]

So, ∠5 = ∠3 = 40o

∠3 and ∠4                             [linear pair]

∠3 + ∠4 = 180o

∠4 = 180o – 40o = 140o

Now, ∠4 and ∠6                [interior angles]

So, ∠4 = ∠6 = 140o

∠3 and ∠7                            [corresponding angles]

So, ∠3 = ∠7 = 40o

Therefore, ∠7 = 40o

∠4 and ∠8                            [corresponding angles]

So, ∠4 = ∠8 = 140o

Therefore, ∠8 = 140o

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

Question :6. In Fig.63, line l || m and a transversal n cuts them P and Q respectively. If ∠1 = 75°, find all other angles.

Solution 6:

In the question given that, l ∥ m and ∠1 = 75o

∠1 = ∠3                                 [vertically opposite angles]

According to the figure

∠1 + ∠2 = 180o    [linear pair]

∠2 = 180o – 75o = 105o

∠1 = ∠5 = 75o                      [corresponding angles]

∠5 = ∠7 = 75o                      [vertically opposite angles]

∠2 = ∠6 = 105o    [corresponding angles]

∠6 = ∠8 = 105o    [vertically opposite angles]

∠2 = ∠4 = 105o    [vertically opposite angles]

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

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

Solution 7:

In the question given that, AB ∥ CD and ∠QMD = 100o

According to the figure ∠QMD + ∠QMC = 180o    [linear pair]

∠QMC = 180o – ∠QMD

∠QMC = 180o – 100° = 80o

Corresponding angles are

∠DMQ = ∠BLM = 100o

∠CMQ = ∠ALM = 80o

Vertically Opposite angles are

∠DMQ = ∠CML = 100o

∠BLM = ∠PLA = 100o

∠CMQ = ∠DML = 80o

∠ALM = ∠PLB = 80o

Question :8. In Fig. 65, l ∥ m and p ∥ q. Find the values of x, y, z, t.

Solution 8:

According to the given information one of the angle is 80o

∠z and 80o           [vertically opposite angles]

Thus ∠z = 80o

∠z and ∠t             [corresponding angles]

∠z = ∠t

Thus, ∠t = 80o

∠z and ∠y             [corresponding angles]

∠z = ∠y

Thus, ∠y = 80o

∠x and ∠y            [corresponding angles]

∠y = ∠x

Thus, ∠x = 80o

Question :9. In Fig. 66, line l ∥ m, ∠1 = 120o and ∠2 = 100o, find out ∠3 and ∠4.

Solution 9:

In the question given that, ∠1 = 120o and ∠2 = 100o

According to the figure ∠1 and ∠5             [linear pair]

∠1 + ∠5 = 180o

∠5 = 180o – 120o = 60o

Thus, ∠5 = 60o

∠2 and ∠6                            [corresponding angles]

∠2 = ∠6 = 100o

Thus, ∠6 = 100o

∠6 and ∠3                            [linear pair]

∠6 + ∠3 = 180o

∠3 = 180o – 100o =80o

Thus, ∠3 = 80o

By, angles of sum property

∠3 + ∠5 + ∠4 = 180o

∠4 = 180o – 80o – 60o =  40o

Thus, ∠4 = 40o

Question :10. In Fig. 67, l ∥ m. Find the values of a, b, c, d. Give reasons.

Solution 10:

According to the question l ∥ m

According to the figure,

∠a = 110o                                             [vertically opposite angles]

∠a = ∠b                 [Corresponding angles]

Thus, ∠b = 110o

∠d = 85o                                               [Vertically opposite angle]

∠d = ∠c                 [Corresponding angles]

Thus, ∠c = 85o

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

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

Solution 11:

According to the given information, ∠1 and ∠2 are in the ratio 3: 2

Let us take the angles as 3x, 2x

∠1 and ∠2                            [linear pair]

3x + 2x = 180o

5x = 180o

x =    = 36o

Thus, ∠1 = 3x = 3(36) = 108o

∠2 = 2x = 2(36) = 72o

∠1 and ∠5                            [corresponding angles]

Thus, ∠1 = ∠5

Hence, ∠5 = 108o

∠2 and ∠6                            [corresponding angles]

So ∠2 = ∠6

Thus, ∠6 = 72o

∠4 and ∠6                            [alternate pair of angles]

∠4 = ∠6 = 72o

Thus, ∠4 = 72o

∠3 and ∠5                            [alternate pair of angles]

∠3 = ∠5 = 108o

Thus, ∠3 = 108o

∠2 and ∠8                            [alternate exterior of angles]

∠2 = ∠8 = 72o

Thus, ∠8 = 72o

∠1 and ∠7                            [alternate exterior of angles]

∠1 = ∠7 = 108o

Thus, ∠7 = 108o

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

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

Solution 12:

According to the question, l, m and n are parallel lines intersected by transversal p at X, Y and Z

∠4 + 60o = 180o                            [linear pair]

∠4 = 180o – 60o

∠4 = 120o

According to the figure

∠4 and ∠1                                            [corresponding angles]

∠4 = ∠1

Thus, ∠1 = 120o

∠1 and ∠2                                            [corresponding angles]

∠2 = ∠1

Thus, ∠2 = 120o

∠2 and ∠3                            [vertically opposite angles]

∠2 = ∠3

Thus, ∠3 = 1200

Question :13. In Fig. 70, if l ∥ m ∥ n and ∠1 = 60o, find ∠2

Solution 13:

According to the question that l ∥ m ∥ n

According to the figure Corresponding angles are

∠1 = ∠3

∠1 = 60o

Thus, ∠3 = 60o

∠3 and ∠4                            [linear pair]

∠3 + ∠4 = 180o

∠4 = 180o – 60o = 120o

∠2 and ∠4                            [alternate interior angles]

∠4 = ∠2

Thus, ∠2 = 120o

Question :14. In Fig. 71, if AB ∥ CD and CD ∥ EF, find ∠ACE

Solution 14:

According to the question, AB ∥ CD and CD ∥ EF

Sum of the interior angles,

∠CEF + ∠ECD = 180o

130o + ∠ECD = 180o

∠ECD = 180o – 130o = 50o

As we know that alternate angles are equal

∠BAC = ∠ACD

∠BAC = ∠ECD + ∠ACE

∠ACE = 70o – 50o

∠ACE = 20o

Thus, ∠ACE = 20o

Question :15. In Fig. 72, if l ∥ m, n ∥ p and ∠1 = 85o, find ∠2.

Solution 15:

According to the question, ∠1 = 85o

∠1 and ∠3                            [corresponding angles]

So, ∠1 = ∠3

∠3 = 85o

Sum of the interior angles is 180o

∠3 + ∠2 = 180o

∠2 = 180o – 85o

∠2 = 95o

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

Solution 16:

According to the question ∠1 = 70o and ∠7 = 80o

As we know, if the alternate exterior angles of the two lines are equal, then the lines are parallel.

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

∠1 ≠ ∠7

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

Solution 17:

According to the figure ∠2 = ∠4 [vertically opposite angles]

∠2 = ∠4 = 65o

∠8 = ∠6 = 65o

Thus, ∠4 = ∠6

So, l ∥ m

Question :18. In Fig. 75, Show that AB ∥ EF.

Solution 18:

As we know,

∠ACD = ∠ACE + ∠ECD

∠ACD = 22o + 35o

∠ACD = 57o = ∠BAC

Thus, lines BA and CD are intersected by line AC

such that, ∠ACD = ∠BAC

So, the alternate angles are equal

Therefore, AB ∥ CD ………….…1

∠ECD + ∠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

Thus, EF ∥ CD ………..….2

Since, equation 1 and 2

AB ∥ EF

Question :19. In Fig. 76, AB ∥ CD. Find the values of x, y, z.

Solution 19:

According to the question that AB ∥ CD

∠x + 125o = 180o                                                 [Linear pair,]

∠x = 180o – 125o

∠x = 55o

∠z = 125o                                                                     [Corresponding angles]

∠x + ∠z = 180o                                    [Adjacent interior angles]

∠x + 125o = 180o

∠x = 180o – 125o

∠x = 55o

∠x + ∠y = 180o

∠y + 55o = 180o

∠y = 180o – 55o

∠y = 125o

Question :20. In Fig. 77, find out ∠PXR, if PQ ∥ RS.

Solution 20:

According to the question, PQ ∥ RS

Find ∠PXR

∠XRS = 50o

∠XPQ = 70o

In the question given that PQ ∥ RS

∠PXR = ∠XRS + ∠XPR

∠PXR = 50o + 70o

∠PXR = 120o

Thus, ∠PXR = 120o

Question :21. In Figure, we have

(i) ∠MLY = 2∠LMQ

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

(iii) ∠XLM = ∠PML, find ∠ALY

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

Solution 21:

(i) ∠MLY = 2∠LMQ

∠MLY and ∠LMQ                               [interior angles]

∠MLY + ∠LMQ = 180o

2∠LMQ + ∠LMQ = 180o

3∠LMQ = 180o

∠LMQ = 180/3 = 60o

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

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

∠XLM and ∠LMQ                               [alternate interior angles]

∠XLM = ∠LMQ

(2x – 10)o = (x + 30)o

2x – x = 30o + 10o = 40o

Thus, x = 40°

(iii) ∠XLM = ∠PML, find ∠ALY

∠XLM = ∠PML

Sum of interior angles is 180 degrees

∠XLM + ∠PML = 180o

∠XLM + ∠XLM = 180o

2∠XLM = 180o

∠XLM = 180/2

∠XLM = 90o

∠XLM and ∠ALY                                 [vertically opposite angles]

Thus, ∠ALY = 90o

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

∠ALY and ∠LMQ                                [corresponding angles]

∠ALY = ∠LMQ

(2x – 15)= (x + 40)o

2x – x = 40o + 15o = 55o

Thus, x = 55o

Question :22. In Fig. 79, DE ∥ BC. Find the values of x and y.

Solution 22:

As we know, ABC, DAB                  [alternate interior angles]

∠ABC = ∠DAB

So, x = 40o

And ACB, EAC                                                    [alternate interior angles]

∠ACB = ∠EAC

So, y = 55o

Question :23. In Fig. 80, line AC ∥ line DE and ∠ABD = 32o, Find out the angles x and y if ∠E = 122o.

Solution 23:

According to the question, line AC ∥ line DE and ∠ABD = 32o

∠BDE = ∠ABD = 32o                          [Alternate interior angles]

∠BDE + y = 180o                 [linear pair]

32+ y = 180o

y = 180o – 32o = 148o

∠ABE = ∠E = 122o                              [Alternate interior angles]

∠ABD + ∠DBE = 122o

32o + x = 122o

x = 122o – 32o =  90o

Question :24. In Fig. 81, side BC of ΔABC has been produced to D and CE ∥ BA. If ∠ABC = 65o, ∠BAC = 55o, find ∠ACE, ∠ECD, ∠ACD.

Solution 24:

∠ABC = 65o, ∠BAC = 55o

∠ABC = ∠ECD = 65o                                        [Corresponding angles]

∠BAC = ∠ACE = 55o                                        [Alternate interior angles]

Now, ∠ACD = ∠ACE + ∠ECD

∠ACD = 55o + 65o

∠ACD = 120o

Question :25. In Fig. 82, line CA ⊥ AB ∥ line CR and line PR ∥ line BD. Find ∠x, ∠y, ∠z.

Solution 25:

In the question given that, CA ⊥ AB

∠CAB = 90o                          ∠AQP = 20o

By, angle of sum property

In ΔABC

∠CAB + ∠AQP + ∠APQ = 180o

∠APQ = 180o – 90o – 20o

∠APQ = 70o

y and ∠APQ                        [corresponding angles]

y = ∠APQ = 70o

∠APQ and ∠z                      [interior angles]

∠APQ + ∠z = 180o

∠z = 180o – 70o = 110o

Question :26. In Fig. 83, PQ ∥ RS. Find the value of x.

Solution 26:

In the question,

∠RCD + ∠RCB = 180o                         [linear pair]

∠RCB = 180o – 130o = 50o

In ΔABC,

∠BAC + ∠ABC + ∠BCA = 180o

By, angle sum property

∠BAC = 180o – 55o – 50o

∠BAC = 75o

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

Solution 27:

∠BAC = ∠ACG = 120o                       [Alternate interior angle]

∠ACF + ∠FCG = 120o

So, ∠ACF = 120o – 90o

= 30o

∠DCA + ∠ACG = 180o                                   [Linear pair]

∠x = 180o – 120o

= 60o

∠BAC + ∠BAE + ∠EAC = 360o

∠CAE = 360o – 120o – (60o + 30o)

= 150o

Question :28. In Fig. 85, AB ∥ CD and AC ∥ BD. Find the values of x, y, z.

Solution 28:

(i)  Since, AC ∥ BD and CD ∥ AB, ABCD                     [parallelogram]

∠CAB + ∠ACD = 180o

∠ACD = 180o – 65o = 115o

Opposite angles of parallelogram,

∠CAB = ∠CDB = 65o

∠ACD = ∠DBA = 115o

(ii)  Since, AC ∥ BD and CD ∥ AB

Alternate interior angles,

∠DAB = y = 35o

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

Solution 29:

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

Here, ∠ACD and ∠CDE                    [alternate and equal angles]

So, ∠ACD = ∠CDE = 100o

Thus, AC ∥ EF

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

Solution 30:

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

Since, AB ∥ DE and BC ∥ EF

The corresponding angles are,

∠ABC = ∠DGC = ∠DEF = 75o