Lesson: Lines and Angles
Exercise 6.1
Question: 1
In Fig. 6.1, if
AB CD EF, PQ RS, RQD 25
and
CQP 60 ,
then
QRS
is equal to:
(a)
85
(b)
135
(c)
145
(d)
110
Solution:
c
Given that,
PQ RS
PQC BRS 60
[Alternate exterior angles and
PQC 60
]
and
[Alternate interior angles
and
DQR 25
]
QRS QRA ARS QRA ( )180 BRS
[Linear pair]
25 180 60
205 60
145
Hence, option (c) is correct.
Question: 2
If one angle of a triangle is equal to the sum of the
other two angles, then the triangle is:
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:
d
Question: 3
An exterior angle of a triangle is
105
and its two
interior opposite angles are equal. Each of these equal
angles is:
(a)
1
37
2
(b)
1
52
2
(c)
1
72
2
(d)
75
Solution:
b
Given,
Exterior angle
105
Let one of interior angle be
x
.
Sum of two opposite interior angles
Exterior
angle
105
x x
2 105
x
105
2
x
1
52
2
x
Hence, each equal angle of triangle is
1
52
2
Question 4
The angles of a triangle are in the ratio 5 : 3 : 7. The
triangle is:
(a) an acute angled triangle
(b) an obtuse angled triangle
(c) a right triangle
(d) an isosceles triangle
Solution:
a
Given that,
The ratio of angles of a triangle is 5 : 3 : 7.
Let the angles of a triangle be
A B ,
and
C
.
Let,
, A 5 B 3 x x
and
C 7 x
In
ABC A B C 180
,
The sum of the angles of a triangle is
180
.
5 3 7 180  x x x
15 180
 x
12
x
5 5 12 60
A x
B 3 3 12 36
x
and
7 7 12 84
C x
Since all the angles are less than
°90
,the triangle is an
acute angled triangle.
Question: 5
If one of the angles of a triangle is 130°, then the angle
formed when the bisectors of the other two angles
meet can be:
(a)
50
(b)
65
(c)
145
(d)
155
Solution:
d
Let the angles of
ABC
be
, A B
and
C
.
In
,ABC
A B C 180
The sum of the angles of a triangle is
180
.
1 1 1 180
A B C
2 2 2 2
Dividing both the sides by 2
1 1 180 1
2 2 2 2
B C A
1 1 1
B C 90 A
2 2 2
Now, BC and OC are the angle bisectors of
ABC
and
ACB
Hence,
1
OBC B
2
.
And
1
BCO C
2
.
In
, OBC OBC BCO COB 180
+
1 1
B C 180 COB
2 2
1
90 A 180 COB
2
(From equation i)
1
COB 90 A
2
COB 90 65
COB 155
Question: 6
In Fig. 6.2, POQ is a line. The value of x is:
(a)
20
(b)
25
(c)
30
(d)
35
Solution:
a
Given,
POQ is a line segment.
Hence,
POQ 180
POA AOB BOQ 180
Putting
, POA 40 AOB 4
x ,
and
BOQ 3 x
40 4 3 180
x x
7 140
x
7 140
x
20
x
Question 7
In Fig. 6.3, if
OP RS OPQ 110
,
and
QRS 130
,
then
PQR
is equal to:
(a)
40
(b)
50
(c)
60
(d)
70
Solution:
c
producing OP such that it intersects RO at point x.
Now,
OP RS
and RX is transversal.
Hence,
RXP QRS
[The alternate interior angles are equal.]
RXP 130
… (i)
[ ]QRS 130
Now, RO is a line segment.
So,
PXQ RXP 180
PXQ 180 RXP
PXQ 180 130
[From eqn (i)]
PXQ 50
In
, PQX OPQ
is an exterior angle.
Hence,
OPQ PXQ PQX
[∴ Exterior angle = sum of two opposite interior
angles.]
110 50 PQX
PQX 60
Question 8
Angles of a triangle are in the ratio 2 : 4 : 3.
The smallest angle of the triangle is
(a)
60
(b)
40
(c)
80
(d)
20
Solution:
b
Given that,
The ratio of angles of a triangle is 2 : 4 : 3.
Let angles of a triangle be
A B ,
and
C
.
, A 2 B 4 x x
and
C 3 x
In
ABC A B C 180
,
[The sum of the angles of a triangle is
180
]
2 4 3 180 x x x
9 180
x
20
x
A 2 2 20 40
x
B 4 4 20 80
x
and
C 3 3 20 60
x
Hence, option (b) is correct.
Exercise 6.2
Question 1
For what value of x y in Fig. 6.4 will ABC be a line?
Justify your answer.
Solution:
For ABC to be a line, the sum of the two adjacent
angles must be
180
or
ABD
and
DBC
must form
a linear pair. i.e.,
180
x y
.
Question 2
Can a triangle have all the angles less than
60
? Give
reason for your answer.
Solution:
No, a triangle cannot have all angles less than
60
,
because if all angles are less than
60
, then their sum
will be less than
180
(not equal to
180
).
Hence, it will not be a triangle.
Question: 3
Can a triangle have two obtuse angles? Give reason
for your answer.
Solution:
No, because if a triangle has two obtuse angles i.e.,
more than
90
angle, then the sum of all three angles
of a triangle will be greater than
180
(not equal to
180
).
Hence, it will not be a triangle.
Question: 4
How many triangles can be drawn with angles
measuring as
45
,
64
and
72
? Give reason for your
answer.
Solution:
None.
The sum of given angles
45 64 72 181 180
.
Hence, we see that sum of all three angles is not equal
to
180
. So, we cannot draw a triangle with the given
angles.
Question: 5
How many triangles can be drawn with angles as
53
,
64
and
63
? Give reason for your answer.
Solution:
The sum of the given angles
53 64 63 180
.
Since, the sum of all interior angles of the triangle is
180
, infinitely many triangles can be drawn.
Question: 6
In Fig. 6.5, find the value of x for which the lines l and
m are parallel.
Solution:
In the given figure,
l m
.
Using the properties of parallel line (if a transversal
intersects two parallel lines, then the sum of interior
angles on the same side of a transversal is
supplementary), we have:
44 180
x
180 44
x
136
x
Question: 7
Two adjacent angles are equal. Is it necessary that
each of these angles will be a right angle? Justify your
answer.
Solution:
No, because each of these will be a right angle only
when they form a linear pair.
Question: 8
If one of the angles formed by two intersecting lines is
a right angle, what can you say about the other three
angles? Give reason for your answer.
Solution:
Let two intersecting lines be l and m. As one of the
angles is a right angle, then it means that lines l and m
are perpendicular to each other. By using linear pair
axiom, other three angles will be right angles.
Question: 9
In Fig. 6.6, which of the two lines are parallel and why?
Solution:
In case (i),
The sum of two interior angles
132 48 180
.
Here, the sum of two interior angles on the same side
of line n is
180
. Hence, l and m are the parallel lines.
In case (ii)
The sum of two interior angles
73 106 179 180
.
Here, the sum of two interior angles on the same side
of the transversal is not equal to
180
. Hence, lines p
and q are not the parallel lines.
Question: 10
Two lines l and m are perpendicular to the same line n.
Are l and m perpendicular to each other? Give reason
for your answer.
Solution:
No.
Let lines l and m be two lines which are perpendicular
to the line n.
1 2 90 90 180
[
l n
and
m n
]
Exercise: 6.3
Question: 1
In Fig. 6.9, OD is the bisector of
AOC
, OE is the
bisector of
BOC
and
OD OE
. Show that the points
A, O and B are collinear.
Solution:
Given:
In figure,
OD OE
and OD is the bisector
of
AOC
and OE is the bisector of
BOC
.
To prove:
Points A, O and B are collinear i.e., AOB is a straight
line.
Proof:
OD and OE bisect angles
AOC
and
BOC
respectively.
AOC=2 DOC
…….. (i) and
BOC=2 COE
…….. (ii)
On adding equations (i) and (ii), we get
AOC+ COB=2 DOC+2 COE
AOC+ COB=2( DOC+ COE)
AOC+ COB=2 DOE
AOC+ COB=2 90 [ OD OE]
AOC+ COB=180
AOB=180
So,
AOC
and
COB
are linear pairs or AOB is a
straight line.
Hence, points A, O and B are collinear.
Question: 2
In Fig. 6.10,
1= 60
and
6 = 120
.Show that the lines
m and n are parallel.
Solution:
Given:
In figure,
1= 60
and
6 = 120
To prove:
m n
.
Proof:
Since,
1= 60
and
6 = 120
Here,
1= 3
[Vertically opposite angles]
3= 1= 60
Now,
3+ 6 = 60 120
3+ 6 = 180
The sum of two interior angles on the same side of the
transversal is
180
.
Hence, the lines are parallel or
m n
.
Question: 3
AP and BQ are the bisectors of the two alternate
interior angles formed by the intersection of a
transversal t with parallel lines l and m (Fig. 6.11).
Show that
AP BQ
.
Solution:
Given:
In figure,
l m
, AP and BQ are the bisectors of
EAB
and
ABH
respectively.
To prove:
AP BQ
Proof:
Since,
l m
and t is transversal,
therefore,
EAB = ABH
[Alternate interior angles]
1 1
EAB = ABH
2 2
[Dividing both sides by 2]
PAB = ABQ
[AP and BQ are the bisectors of
EAB
and
ABH
]
Since,
PAB
and
ABQ
are alternate interior angles
formed when transversal AB intersects lines AP and
BQ, hence,
AP BQ
.
Question: 4
If in Fig. 6.11, bisectors AP and BQ of the alternate
interior angles are parallel, then show that
l m
.
Solution:
Given:
In figure,
AP BQ
, AP and BQ are the bisectors of
alternate interior angles
EAB
and
ABH
.
To prove:
l m
Proof:
Since,
AP BQ
and t is the transversal, therefore
PAB = ABQ
[Alternate interior angles].
2 PAB =2 ABQ 
[Multiplying both sides by 2]
EAB = ABH
As the alternate interior angles
EAB
and
ABH
are
equal, the lines l and m are parallel. [If two alternate
interior angles are equal, then the lines are parallel.]
Question: 5
In Fig. 6.12,
BA ED
and
BC EF
. Show that
ABC = DEF
[Hint: Produce DE to intersect BC at P (say)].
Solution:
Given:
BA ED
, and
BC EF
.
To prove:
ABC = DEF
Construction:
Produce DE which intersects BC at P.
Proof:
In figure,
BA ED BA DP
ABP= EPC
[Corresponding angles]
ABC = EPC
… (i)
Again,
BC EF PC EF
EPC = DEF
… (ii)
[Corresponding angles]
From equations (i) and (ii), we get
ABC= DEF
Hence, it is proved.
Question: 6
In Fig. 6.13,
BA ED
and
BC EF
.
Show that
ABC+ DEF=180
.
Solution:
Given:
In the figure,
BA ED
and
BC EF
.
To prove:
ABC+ DEF=180
.
Construction:
Draw a ray PE opposite to ray EF.
Proof:
In figure,
BC EF
or
BC PF
EPB+ PBC=180
... (i)
[The sum of the interior angles on the same side of the
transversal is 180°.]
Now,
AB ED
and PE is a transversal line.
EPB+ DEF
... (ii) [Corresponding angles]
From equations (i) and (ii), we get
DEF+ PBC=180

ABC+ DEF=180
[
PBC+ ABC
]
Hence, it is proved.
Question: 7
In Fig. 6.14,
DE QR
and AP and BP are the bisectors
of
EAB
and
RBA
, respectively. Find
APB
.
Solution:
Given:
DE QR
, and AP and PB are the bisectors of
EAB
and
RBA
, respectively.
To find:
APB
Proof:
As
DE QR
and AB is a transversal, the sum of the
interior angles on the same side of the transversal is
supplementary.
EAB+ RBA =180
1 1 180
EAB + RBA =
2 2 2
[Dividing both sides by 2]
1 1
EAB + RBA =90
2 2
… (i)
Since AP and BP are the bisectors of
EAB
and
RBA
respectively,
1
BAP + EAB
2

… (ii)
and
1
ABP+ RBA
2

… (iii)
On adding equations (ii) and (iii), we get
1 1
BAP+ ABP= EAB+ RBA
2 2
Using equation (i), we get
BAP+ ABP=90
… (iv)
In
ABP
,
BAP+ ABP+ APB=180
[The sum of the interior angles in a triangle is
180
.]
90 + APB =180
[From equation (iv)]
APB 180 90 = 90
Hence,
APB 90
Question: 8
The angles of a triangle are in the ratio 2 : 3 : 4. Find
the angles of the triangle.
Solution:
Given,
The ratio of the angles of a triangle is 2 : 3 : 4.
Let the angles of a triangle be
A B ,
and
C
.
, A 2 B 3 x x
and
C 4 x
In
0ABC A B C 18
,
[The sum of the angles of a triangle is
180
]
2 3 4 =180
x x x
9 180
x
20
x
A 2 2 20 40
x
B 3 3 20 60
x
and
C 4 4 20 80
x
Question: 9
A triangle ABC is right angled at A. L is a point on
BC such that
AL BC
. Prove that
BAL ACB
.
Solution:
Given:
In
ABC, A 90
and
AL BC
To prove:
BAL ACB
Proof:
In
ABC
and
LAC
,
BAC ALC
… (i)
and
ABC ABL
..… (ii) [Common angle]
Adding equations (i) and (ii), we get
BAC ABC ALC+ ABL
..… (iii)
In
ABC, BAC ACB ABC 180
[The sum of the angles of a triangle is
180
.]
BAC ABC 180 ACB
..… (iv)
In
ABL, ABL ALB BAL 180
[The sum of the angles of a triangle is
180
]
ABL ALC 180 BAL
..… (v)
[
ALC ALB 90
]
Substituting the value from equation (iv) and (v) in
equation (iii), we get
180 ACB 180 BAL
ACB BAL
Hence, it is proved.
Question: 10
Two lines are respectively perpendicular to two
parallel lines. Show that they are parallel to each other.
Solution:
Given:
Two lines m and n are parallel and another two lines p
and q are perpendicular to m and n respectively.
Or
p m
and
p n , q m
and
q n
To prove:
p q
Proof:
Since,
m n
and p is perpendicular to m and n.
So,
p is perpendicular to m …(i)
p is perpendicular to n …(ii)
Since, m ∥ n and q is perpendicular to m and n.
So,
q is perpendicular to m …(iii)
q is perpendicular to n …(iv)
From the equations (i) and (iii) [Or from (ii) and (iv)],
we get,
p q
.
[If two lines are perpendicular to the same line, the
lines are parallel to each other.]
Hence, it is
p q
.
Exercise: 6.4
Question: 1
If two lines intersect, prove that the vertically opposite
angles are equal.
Solution:
Given:
Two lines AB and CD intersect at point O.
To prove:
(i)
AOC BOD
(ii)
AOD BOC
Proof:
(i) Since, ray OA stands on line CD,
AOC AOD 180
… (i)
[Linear pair axiom]
Similarly, ray OD stands on line AB.
AOD BOD 180
… (ii)
From equations (i) and (ii), we get
AOC AOD AOD BOD
AOC BOD
(ii) Since, ray OD stands on line AB,
AOD BOD 180
… (iii)
[Linear pair axiom]
Similarly, ray OB stands on line CD.
DOB BOC 180
… (iv)
From equations (iii) and (iv), we get
AOD BOD DOB BOC
AOD BOC
Question: 2
Bisectors of interior
B
and exterior
ACD
of
ABC
intersect at the point T.
Prove that:
1
BTC BAC
2
.
Solution:
Given:
In
ABC
, the bisectors of
ABC
and
ACD
meet at
point T.
Construction:
Produce BC to D.
To Prove:
1
BTC BAC
2
Proof:
In
ABC, ABC
is an exterior angle.
ACD, ABC+ CAB
[The exterior angle of a triangle is equal to the sum of
two opposite interior angles.]
1 1 1
ACD CAB+ ABC
2 2 2
[Dividing both sides by 2]
1 1
TCD CAB+ ABC
2 2
… (i)
[∵ CT is a bisector of
1
ACD ACD TCD
2
]
In
BTC
,
TCD BTC+ CBT
[The exterior angle of a triangle is equal to the sum of
two interior opposite angles.]
1
TCD BTC+ ABC
2
… (ii)
[∵ BT bisector of
1
ABC CBT ABC
2
]
From equations (i) and (ii), we get
1 1 1
CAB+ ABC BTC ABC
2 2 2
1
CAB BTC
2
or
1
BAC BTC
2
Hence, it is proved.
Question: 3
A transversal intersects two parallel lines. Prove that
the bisectors of any pair of corresponding angles so
formed are parallel.
Solution:
Given:
Two lines DE and QR are parallel. A transversal
intersects the lines DE and QR at A and B respectively.
Also, BP and AF are the bisectors of angles
ABR
and
CAE
respectively.
To prove:
BP AF
Proof:
Given,
DE QR
CAE ABR
[Corresponding angles]
1 1
CAE ABR
2 2
[Dividing both sides by 2]
CAF ABP
[∵ BP and AF are the bisectors of angles
ABR
and
CAE
respectively]
This implies that AF is parallel to BP as the
corresponding angles
CAF
and
ABP
are equal.
Hence,
AF BP
.
Question: 4
Prove that through a given point, we can draw only
one perpendicular on a given line.
[Hint: Use proof by contradiction].
Solution:
Given:
Consider a line l and a point P.
To prove:
Only one perpendicular can be drawn from P to l.
Construction:
Suppose that two lines PA and PB are passing through
the point P and they are perpendicular to l.
Proof:
In
APB
,
PAB P PBA 180
[The angle sum property of a triangle]
90 P 90 180
(since PA is perpendicular to
l and PB is perpendicular to l)
P 180 180
P 0
(Which is possible only when the lines PA and PB
coincide)
Hence, only one perpendicular line can be drawn
through a given point.
Question: 5
Prove that two lines that are respectively
perpendicular to two intersecting lines intersect each
other.
[Hint: Use proof by contradiction].
Solution:
Given:
Let lines l and m be two intersecting lines. Again, let n
and p be another two lines which are perpendicular to
the intersecting lines.
To prove:
Two lines n and p intersect at a point.
Proof:
Let us consider lines n and p are not intersecting, and
then it means they are parallel to each other i.e.,
n p
… (i)
Since, lines n and p are perpendicular to l and m
respectively, but from equation (i)
n p
, it is implied
that
l m
, it is a contradiction.
Thus, our assumption is wrong.
Hence, lines n and p intersect at a point.
Question: 6
Prove that a triangle must have at least two acute
angles.
Solution:
Given:
ABC
is a triangle
To prove:
ABC
must have two acute angles.
Proof:
Let us consider the following cases:
Case I:
When two angles are
90
:
Suppose two angles are
B 90
and
C 90
.
The sum of all three angles of a triangle is
180
.
A B C 180
A 90 90 180
A 180 180 0
, which is not possible.
Hence, this case is rejected.
Case II:
When two angle are obtuse:
Suppose two angles
B
and
C
are more than
90
,
the sum of all three angles of a triangle is
180
.
A B C 180
A 180 ( B C) 180
− (an angle greater
than
180
)
A
= negative angle, which is not possible.
Hence, this case is also rejected.
Case III:
When one angle is
90
and the other is obtuse:
Suppose
B 90
and
C
is obtuse.
The sum of all three angles of a triangle is
180
.
A B C 180
A 180 (90 C)
90 C
= negative angle, which is not possible.
Hence, this case is also rejected.
Case IV:
When two angles are acute, then the sum of two
angles is less than
180
so that the third angle is also
acute.
Hence, a triangle must have at least two acute angles.
Question: 7
In Fig. 6.17,
Q R, PA
is the bisector of
QPS
and
PM QR
. Prove that
1
APM Q R
2
.
Solution:
Given:
Q R, PA
is the bisector of
QPR
and
PM QR
.
To prove:
1
APM Q R
2
.
Proof:
Since, PA is the bisector of
QPR
QPA APR
… (i)
In
PQM, Q PMQ QPM 180
[ngle sum property of triangles]
Q 90 QPM 180
[
PMQ 90
]
Q 90 QPM
… (ii)
In
PMR, PMR R RPM 180
[Angle sum property of triangles]
90 R RPM 180
[
PMQ 90
]
R 180 90 RPM
R 90 RPM
… (iii)
Subtracting equations (iii) from (ii), we get
Q R (90 QPM) (90 RPM)
Q R RPM QPM
Q R ( RPA APM) ( QPA APM)  
… (iv)
Q R QPA APM QPA APM
[sing equation (i)]
Q R 2 APM
1
APM ( Q R)
2