**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

∠BAC and ∠CAD

In fig (ii), the adjacent angles are

∠BAD and ∠DAC

∠BDA and ∠CDA

**Question :3. In fig. 15, write down**

**(i) Each linear pair**

**(ii) Each pair of vertically opposite angles.**

**Solution 3:**

**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) 35 ^{o}**

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

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

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

**Solution 5:**

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

(i) Thus, complementary angle for given angle is

90^{o} – 35^{o} = 55^{o}

(ii) Thus, complementary angle for given angle is

90^{0} – 72^{o} = 18^{o}

(iii) Thus, complementary angle for given angle is

90^{o }– 45^{o} = 45^{o}

(iv) Thus, complementary angle for given angle is

90^{o} – 85^{o} = 5^{o}

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

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

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

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

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

**Solution 6:**

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

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

180^{o} – 70^{o} = 110^{o}

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

180^{o} – 120^{o} = 60^{o}

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

180^{o} – 135^{o} = 45^{o}

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

180^{o} – 90^{o} = 90^{o}

** **

**Question :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 7:**

(i) 25^{o} + 65^{o} = 90^{o}

So, this is a pair of complementary angles.

(ii) 120^{o} + 60^{o} = 180^{o}

So, this is a pair of supplementary angles.

(iii) 63^{o} + 27^{o} = 90^{o}

So, this is a pair of complementary angles.

(iv) 100^{o} + 80^{o} = 180^{o}

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 180^{o}.

(ii) Yes, two right angles can be supplementary

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

(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 90^{o}

** **

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

**Solution 9:**

If the sum of the two angles is 180^{o}, 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 = 180^{o}

∠p + ∠p = 180^{o}

2∠p = 180^{o}

∠p =

∠p = 90^{o}

Thus, ∠p = ∠q = 90^{o}

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

**Solution 12:**

According to the question complement of an angle is 28^{o}

Let x be the complement of the angle 28^{o}

So, ∠x + 28^{o} = 90^{o}

∠x = 90^{o} – 28^{o} = 62^{o}

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

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

**Solution 14:**

Given in the question, ∠1 = 70^{o}

∠3 = 2(∠1)

= 2(70^{o})

∠3 = 140^{o}

∠3 = ∠4

As, OE is the angle bisector,

∠DOB = 2(∠1)

∠DOB = 2(70^{o})

∠DOB = 140^{o}

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

140^{o} + 140^{o} + 2(∠COB) = 360^{o}

Since, ∠COB = ∠AOD

2(∠COB) = 360^{o} – 280^{o}

2(∠COB) = 80^{o}

∠COB = 80^{o}/2

∠COB = 40^{o}

Thus, ∠COB = ∠AOD = 40^{o}

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

** **

**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 90^{o}

Linear pair angle is 180^{o}

180^{o} – 90^{o} = 90^{o}

So, the other angle is 90^{o}.

** **

**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 180^{o}.

** **

**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 180^{o}.

** **

**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 180^{o}.

** **

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

**Solution 19:**

Let x be the required angle

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

x = 180^{o} – 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.

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

**Solution 20:**

(i) ∠BOA + ∠BOC = 180^{o}

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

x^{o} = 180^{o} – 60^{o}

x^{o} = 120^{o}

(ii) ∠POQ + ∠QOR = 180^{o}

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

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

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

x^{o} = 36^{o}

(iii) ∠LOP + ∠PON + ∠NOM = 180^{o}

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

x^{o} = 180^{o} – 35^{o} – 60^{o}

x^{o} = 180^{o} – 95^{o}

x^{o} = 85^{o}

(iv) ∠DOC + ∠DOE + ∠EOA + ∠AOB+ ∠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} – 297^{o}

x^{o} = 63^{o}

(v) ∠ROS + ∠ROQ + ∠QOP + ∠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) 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 180^{o}

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

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

x^{o} = 35^{o}

** **

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

**Solution 21:**

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

Thus, ∠3 = 65^{o}

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 180^{o}

Thus, ∠2 = 180^{o} – 65^{o} = 115^{o}

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

Therefore, ∠2 = ∠4 = 115^{o}

And ∠1 = 65^{o}

** **

**Question :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 22:**

(i) ∠AOC + ∠BOC = 180^{o}

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 180^{o}

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

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

3x + 2y = 175^{o}

Given in the question, If x = 25^{o}, then

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

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

2y = 175^{o} – 75^{o}

2y = 100^{o}

y = = 50^{o}

(ii) ∠AOC + ∠BOC = 180^{o}

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 180^{o}

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

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

3x + 2y = 175^{o}

Given in the question, If y = 35^{o}, then

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

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

3x = 175^{0} – 70^{o}

3x = 105^{o}

x =

x = 35^{o}

** **

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

**Solution 23:**

Pairs of adjacent angles are:

∠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 180^{o}

∠AOD and ∠BOD ∠AOC and ∠BOC

** **

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

**Solution 24:**

(x + 10)^{ o} + x^{o} + (x + 20)^{o} = 180^{o}[linear pair]

By solving the equation

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

3x^{o} = 180^{o} – 30^{o}

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

x^{o} =

x^{o} = 50^{o}

Also, ∠BOC = (x + 20)^{o }[x=50]

∠BOC = (50 + 20)^{o} = 70^{o}

∠COD = 50^{o}

∠AOD = (x + 10)^{o}

∠AOD = (50 + 10)^{o} = 60^{o}

** **

**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^{o }[linear pair]

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

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

x^{o} = = 30^{o}

** **

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

**Solution 28:**

According to the figure we can written,

∠AOB + ∠BOC = 180^{o} [linear pair]

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

2x = 180^{o} – 70^{o}

2x = 110^{o}

x = = 55^{o}

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

**Solution 29:**

According to the figure we can written

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

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

By solving the equation

100^{o} + 4x = 180^{o}

4x = 180^{o} – 100^{o}

4x = 80^{o}

x = = 20^{o}

** **

**Question :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 30:**

In the question given that, ∠x = 45^{o}

According to the figure we can written

∠x = ∠z = 45^{o}

Also, we have

∠y = ∠u

Since the property of linear pair,

=∠x + ∠y + ∠z + ∠u = 360^{o}

=45^{o} + 45^{o} + ∠y + ∠u = 360^{o }

=90^{o} + ∠y + ∠u = 360^{o}

=∠y + ∠u = 360^{o} – 90^{o}

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

=2∠y = 270^{o}

=∠y =

=∠y = 135^{o}

∠y = ∠u = 135^{o}

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

** **

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

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

∠x + 140^{o} = 180^{o}

By solving the equation

∠x = 180^{o} – 140^{o} = 40^{o}

According the figure we written as

∠x = ∠u = 40^{o } [vertically opposite angles]

∠z = 90^{o } [vertically opposite angle]

∠y = 50^{o } [vertically opposite angle]

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

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

**Solution 32:**

∠y = 25^{o} [vertically opposite angle]

According to the figure we written as

∠x = ∠z [vertically opposite angles]

∠x + ∠y + ∠z + 25^{o} = 360^{o}

∠x + ∠z + 25^{o} + 25^{o} = 360^{o}

By solving the equation

∠x + ∠z + 50^{o} = 360^{o}

∠x + ∠z = 360^{o} – 50^{o} [∠x = ∠z]

2∠x = 310^{o}

∠x = 155^{o}

And, ∠x = ∠z = 155^{o}

Thus, ∠x = 155^{o}, ∠y = 25^{o} and ∠z = 155^{o}

**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 = 60 ^{o}, 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 = 60^{o} [given in question]

Vertically opposite angles ∠LMD = ∠CMQ = 60^{o }[given in question]

Vertically opposite angles ∠ALM = ∠PLB = 60^{o}

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

By solving the equation

∠QMD = 180^{o} – 60^{o} = 120^{o}

Corresponding angles are

∠QMD = ∠MLB = 120^{o}

Vertically opposite angles

∠QMD = ∠CML = 120^{o}

∠MLB = ∠ALP = 120^{o}

** **

**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 = 35 ^{o} find ∠ALM and ∠PLA.**

**Solution 3:**

In the question given that, ∠LMD = 35^{o}

According to the figure we written

∠LMD and ∠LMC [linear pair]

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

By solving the equation

∠LMC = 180^{o} – 35^{o} = 145^{o}

So, ∠LMC = ∠PLA = 145^{o}

And, ∠LMC = ∠MLB = 145^{o}

∠MLB and ∠ALM [linear pair]

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

∠ALM = 180^{o} – 145^{o} = 35^{0}

Thus, ∠ALM = 35^{o}, ∠PLA = 145^{o}.

** **

**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 = 40^{o}

∠1 and ∠2 [linear pair]

∠1 + ∠2 = 180^{o}

∠2 = 180^{o} – 40^{o}

∠2 = 140^{o}

Again, from the figure:

∠2 and ∠6 [corresponding angle pair]

So, ∠6 = 140^{o}

∠6 and ∠5 [linear pair]

∠6 + ∠5 = 180^{o}

∠5 = 180^{o} – 140^{o} = 40^{o}

According to the figure we written as

∠3 and ∠5 [alternate interior angles]

So, ∠5 = ∠3 = 40^{o}

∠3 and ∠4 [linear pair]

∠3 + ∠4 = 180^{o}

∠4 = 180^{o} – 40^{o} = 140^{o}

Now, ∠4 and ∠6 [interior angles]

So, ∠4 = ∠6 = 140^{o}

∠3 and ∠7 [corresponding angles]

So, ∠3 = ∠7 = 40^{o}

Therefore, ∠7 = 40^{o}

∠4 and ∠8 [corresponding angles]

So, ∠4 = ∠8 = 140^{o}

Therefore, ∠8 = 140^{o}

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

**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 = 75^{o}

∠1 = ∠3 [vertically opposite angles]

According to the figure

∠1 + ∠2 = 180^{o} [linear pair]

∠2 = 180^{o} – 75^{o} = 105^{o}

∠1 = ∠5 = 75^{o} [corresponding angles]

∠5 = ∠7 = 75^{o} [vertically opposite angles]

∠2 = ∠6 = 105^{o} [corresponding angles]

∠6 = ∠8 = 105^{o} [vertically opposite angles]

∠2 = ∠4 = 105^{o} [vertically opposite angles]

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

** **

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

**Solution 7:**

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

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

∠QMC = 180^{o} – ∠QMD

∠QMC = 180^{o} – 100° = 80^{o}

Corresponding angles are

∠DMQ = ∠BLM = 100^{o}

∠CMQ = ∠ALM = 80^{o}

Vertically Opposite angles are

∠DMQ = ∠CML = 100^{o}

∠BLM = ∠PLA = 100^{o}

∠CMQ = ∠DML = 80^{o}

∠ALM = ∠PLB = 80^{o}

** **

**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 80^{o}

∠z and 80^{o} [vertically opposite angles]

Thus ∠z = 80^{o}

∠z and ∠t [corresponding angles]

∠z = ∠t

Thus, ∠t = 80^{o}

∠z and ∠y [corresponding angles]

∠z = ∠y

Thus, ∠y = 80^{o}

∠x and ∠y [corresponding angles]

∠y = ∠x

Thus, ∠x = 80^{o}

** **

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

**Solution 9:**

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

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

∠1 + ∠5 = 180^{o}

∠5 = 180^{o} – 120^{o} = 60^{o}

Thus, ∠5 = 60^{o}

∠2 and ∠6 [corresponding angles]

∠2 = ∠6 = 100^{o}

Thus, ∠6 = 100^{o}

∠6 and ∠3 [linear pair]

∠6 + ∠3 = 180^{o}

∠3 = 180^{o} – 100^{o} =80^{o}

Thus, ∠3 = 80^{o}

By, angles of sum property

∠3 + ∠5 + ∠4 = 180^{o}

∠4 = 180^{o} – 80^{o} – 60^{o} = 40^{o}

Thus, ∠4 = 40^{o}

** **

**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 = 110^{o }[vertically opposite angles]

∠a = ∠b [Corresponding angles]

Thus, ∠b = 110^{o}

∠d = 85^{o}_{ }[Vertically opposite angle]

∠d = ∠c [Corresponding angles]

Thus, ∠c = 85^{o}

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

** **

**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 = 180^{o}

5x = 180^{o}

x = = 36^{o}

Thus, ∠1 = 3x = 3(36) = 108^{o}

∠2 = 2x = 2(36) = 72^{o}

∠1 and ∠5 [corresponding angles]

Thus, ∠1 = ∠5

Hence, ∠5 = 108^{o}

∠2 and ∠6 [corresponding angles]

So ∠2 = ∠6

Thus, ∠6 = 72^{o}

∠4 and ∠6 [alternate pair of angles]

∠4 = ∠6 = 72^{o}

Thus, ∠4 = 72^{o}

∠3 and ∠5 [alternate pair of angles]

∠3 = ∠5 = 108^{o}

Thus, ∠3 = 108^{o}

∠2 and ∠8 [alternate exterior of angles]

∠2 = ∠8 = 72^{o}

Thus, ∠8 = 72^{o}

∠1 and ∠7 [alternate exterior of angles]

∠1 = ∠7 = 108^{o}

Thus, ∠7 = 108^{o}

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

** **

**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 + 60^{o} = 180^{o }[linear pair]

∠4 = 180^{o} – 60^{o}

∠4 = 120^{o}

According to the figure

∠4 and ∠1 [corresponding angles]

∠4 = ∠1

Thus, ∠1 = 120^{o}

∠1 and ∠2 [corresponding angles]

∠2 = ∠1

Thus, ∠2 = 120^{o}

∠2 and ∠3 [vertically opposite angles]

∠2 = ∠3

Thus, ∠3 = 120^{0}

** **

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

**Solution 13:**

According to the question that l ∥ m ∥ n

According to the figure Corresponding angles are

∠1 = ∠3

∠1 = 60^{o}

Thus, ∠3 = 60^{o}

∠3 and ∠4 [linear pair]

∠3 + ∠4 = 180^{o}

∠4 = 180^{o} – 60^{o} = 120^{o}

∠2 and ∠4 [alternate interior angles]

∠4 = ∠2

Thus, ∠2 = 120^{o}

** **

**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 = 180^{o}

130^{o} + ∠ECD = 180^{o}

∠ECD = 180^{o} – 130^{o} = 50^{o}

As we know that alternate angles are equal

∠BAC = ∠ACD

∠BAC = ∠ECD + ∠ACE

∠ACE = 70^{o} – 50^{o}

∠ACE = 20^{o}

Thus, ∠ACE = 20^{o}

** **

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

**Solution 15:**

According to the question, ∠1 = 85^{o}

∠1 and ∠3 [corresponding angles]

So, ∠1 = ∠3

∠3 = 85^{o}

Sum of the interior angles is 180^{o}

∠3 + ∠2 = 180^{o}

∠2 = 180^{o} – 85^{o}

∠2 = 95^{o}

** **

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

**Solution 16:**

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

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

**Solution 17:**

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

∠2 = ∠4 = 65^{o}

∠8 = ∠6 = 65^{o}

Thus, ∠4 = ∠6

So, l ∥ m

** **

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

**Solution 18:**

As we know,

∠ACD = ∠ACE + ∠ECD

∠ACD = 22^{o} + 35^{o}

∠ACD = 57^{o} = ∠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 = 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

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 + 125^{o} = 180^{o }[Linear pair,]

∠x = 180^{o} – 125^{o}

∠x = 55^{o}

∠z = 125^{o }[Corresponding angles]

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

∠x + 125^{o} = 180^{o}

∠x = 180^{o} – 125^{o}

∠x = 55^{o}

∠x + ∠y = 180^{o}

∠y + 55^{o} = 180^{o}

∠y = 180^{o} – 55^{o}

∠y = 125^{o}

** **

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

**Solution 20:**

According to the question, PQ ∥ RS

Find ∠PXR

∠XRS = 50^{o}

∠XPQ = 70^{o}

In the question given that PQ ∥ RS

∠PXR = ∠XRS + ∠XPR

∠PXR = 50^{o} + 70^{o}

∠PXR = 120^{o}

Thus, ∠PXR = 120^{o}

** **

**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)^{o }, find x.**

**Solution 21:**

(i) ∠MLY = 2∠LMQ

∠MLY and ∠LMQ [interior angles]

∠MLY + ∠LMQ = 180^{o}

2∠LMQ + ∠LMQ = 180^{o}

3∠LMQ = 180^{o}

∠LMQ = 180/3 = 60^{o}

(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 = 30^{o} + 10^{o} = 40^{o}

Thus, x = 40°

(iii) ∠XLM = ∠PML, find ∠ALY

∠XLM = ∠PML

Sum of interior angles is 180 degrees

∠XLM + ∠PML = 180^{o}

∠XLM + ∠XLM = 180^{o}

2∠XLM = 180^{o}

∠XLM = 180/2

∠XLM = 90^{o}

∠XLM and ∠ALY [vertically opposite angles]

Thus, ∠ALY = 90^{o}

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

∠ALY and ∠LMQ [corresponding angles]

∠ALY = ∠LMQ

(2x – 15)^{o }= (x + 40)^{o}

2x – x = 40^{o} + 15^{o} = 55^{o}

Thus, x = 55^{o}

** **

**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 = 40^{o}

And ACB, EAC [alternate interior angles]

∠ACB = ∠EAC

So, y = 55^{o}

** **

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

**Solution 23:**

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

∠BDE = ∠ABD = 32^{o} [Alternate interior angles]

∠BDE + y = 180^{o} [linear pair]

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

y = 180^{o} – 32^{o} = 148^{o}

∠ABE = ∠E = 122^{o} [Alternate interior angles]

∠ABD + ∠DBE = 122^{o}

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

x = 122^{o} – 32^{o} = 90^{o}

** **

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

**Solution 24:**

∠ABC = 65^{o}, ∠BAC = 55^{o}

∠ABC = ∠ECD = 65^{o }[Corresponding angles]

∠BAC = ∠ACE = 55^{o }[Alternate interior angles]

Now, ∠ACD = ∠ACE + ∠ECD

∠ACD = 55^{o} + 65^{o}

∠ACD = 120^{o}

** **

**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 = 90^{o} ∠AQP = 20^{o}

By, angle of sum property

In ΔABC

∠CAB + ∠AQP + ∠APQ = 180^{o}

∠APQ = 180^{o} – 90^{o} – 20^{o}

∠APQ = 70^{o}

y and ∠APQ [corresponding angles]

y = ∠APQ = 70^{o}

∠APQ and ∠z [interior angles]

∠APQ + ∠z = 180^{o}

∠z = 180^{o} – 70^{o} = 110^{o}

** **

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

**Solution 26:**

In the question,

∠RCD + ∠RCB = 180^{o } [linear pair]

∠RCB = 180^{o} – 130^{o} = 50^{o}

In ΔABC,

∠BAC + ∠ABC + ∠BCA = 180^{o}

By, angle sum property

∠BAC = 180^{o} – 55^{o} – 50^{o}

∠BAC = 75^{o}

** **

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

**Solution 27:**

∠BAC = ∠ACG = 120^{o} [Alternate interior angle]

∠ACF + ∠FCG = 120^{o}

So, ∠ACF = 120^{o} – 90^{o}

= 30^{o}

∠DCA + ∠ACG = 180^{o }[Linear pair]

∠x = 180^{o} – 120^{o}

= 60^{o}

∠BAC + ∠BAE + ∠EAC = 360^{o}

∠CAE = 360^{o} – 120^{o} – (60^{o} + 30^{o})

= 150^{o}

** **

**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]

Adjacent angles of parallelogram,

∠CAB + ∠ACD = 180^{o}

∠ACD = 180^{o} – 65^{o} = 115^{o}

Opposite angles of parallelogram,

∠CAB = ∠CDB = 65^{o}

∠ACD = ∠DBA = 115^{o}

(ii) Since, AC ∥ BD and CD ∥ AB

Alternate interior angles,

∠CAD = x = 40^{o}

∠DAB = y = 35^{o}

** **

**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 = 100^{o}

Thus, AC ∥ EF

** **

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