 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?
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
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
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
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
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
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 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
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.
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.
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
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