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

DQR QRA 25

[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

130

COB 90

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 110 50

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