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
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
90o – 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:
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 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 = 180o [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 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 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 = 60o [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 = 180o [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)o , 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)o = (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]
32o + 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]
Adjacent angles of 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,
∠CAD = x = 40o
∠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