# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 8 – Triangles

## Chapter 8 – Triangles Exercise Ex. 8

Question 1

In ABC, if B = 76o and C = 48o, find A.

Solution 1

Since,
sum of the angles of a triangle is 180o

A + B + C = 180o

A + 76o
+ 48o = 180o

A = 180o – 124o = 56o

A = 56o

Question 2

In a right-angled triangle, one of the acute angles measures 53o.
Find the measure of each angle of the triangle.

Solution 2

Let ABC be a right angled triangle and C = 90o

Since,
A + B + C = 180o

A + B = 180oC = 180o – 90o = 90o

Suppose A = 53o

Then, 53o + B = 90o

B = 90o – 53o = 37o

The required angles are 53o, 37o
and 90o.

Question 3

In
a right-angled triangle, one of the acute angles measures 53o.
Find the measure of each angle of the triangle.

Solution 3

Let
ABC be a right angled triangle and C = 90o

Since,
A + B + C = 180o

A + B = 180oC = 180o – 90o = 90o

Suppose A = 53o

Then, 53o + B = 90o

B = 90o – 53o = 37o

The required angles are 53o, 37o
and 90o.

Question 4

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Solution 4

Let ABC be a triangle.

So, A < B + C

Adding A to both sides of the inequality,

A < A + B + C

A < 180o [Since A + B + C = 180o]

Similarly, B <A + C

B < 90o

and C < A + B

C < 90o

ABC is an acute angled triangle.

Question 5

If
one angle of a triangle is greater than the sum of the other two, show that
the triangle is obtuse angled.

Solution 5

Let
ABC be a triangle and B > A + C

Since,
A + B + C = 180o

A + C = 180oB

Therefore, we get,

B > 180o B

B on both sides of the inequality, we get,

B + B > 180oB + B

2B > 180o

B >

i.e.,
B > 90o which means B is an obtuse angle.

ABC is an obtuse angled triangle.

Question 6

In
the given figure, side BC of ABC is produced to D. If ACD = 128o and ABC = 43o, find BAC and ACB.

Solution 6

Since
ACB and ACD form a linear pair.

So, ACB + ACD = 180o

ACB + 128o = 180o

ACB = 180o – 128 = 52o

Also, ABC + ACB + BAC = 180o

43o + 52o
+ BAC = 180o

95o + BAC = 180o

BAC = 180o – 95o = 85o

ACB = 52o and BAC = 85o.

Question 7

In the given figure, the side BC of ABC has been produced on the right-hand side from B to D and  on the right-hand side from C and E. If ABD = 106o and ACE= 118o, find the measure of each angle of the triangle.

Solution 7

As DBA and ABC form a linear pair.

So,DBA + ABC = 180o

106o + ABC = 180o

ABC = 180o – 106o = 74o

Also, ACB and ACE form a linear pair.

So,ACB + ACE = 180o

ACB + 118o = 180o

ACB = 180o – 118o = 62o

In ABC, we have,

ABC + ACB + BAC = 180o

74o + 62o + BAC = 180o

136o + BAC = 180o

BAC = 180o – 136o = 44o

In triangle ABC, A = 44o, B = 74o and C = 62o

Question 8

Calculate
the value of x in each of the following figures.

(i)

(ii)

(iii)

Given:
AB || CD

(vi)

Solution 8

(i) EAB + BAC = 180o [Linear pair angles]

110o + BAC = 180o

BAC = 180o – 110o = 70o

Again, BCA + ACD = 180o [Linear
pair angles]

BCA + 120o
= 180o

BCA = 180o – 120o = 60o

Now,
in ABC,

ABC + BAC + ACB = 180o

xo + 70o
+ 60o = 180o

x + 130o
= 180o

x = 180o – 130o
= 50o

x = 50

(ii)

In
ABC,

A
+ B + C = 180o

30o + 40o
+ C = 180o

70o
+ C = 180o

C = 180o – 70o = 110o

Now BCA + ACD = 180o [Linear pair]

110o
+ ACD = 180o

ACD = 180o – 110o = 70o

In
ECD,

ECD + CDE + CED = 180o

70o + 50o
+ CED = 180o

120o
+ CED = 180o

CED = 180o – 120o = 60o

Since
AED and CED from a linear pair

So, AED + CED = 180o

xo
+ 60o = 180o

xo
= 180o – 60o = 120o

x =
120

(iii)

EAF
= BAC
[Vertically opposite angles]

BAC = 60o

In
ABC, exterior ACD is equal to the sum of two opposite interior
angles.

So, ACD = BAC + ABC

115o = 60o
+ xo

xo = 115o
– 60o = 55o

x = 55

(iv)

Since
AB || CD and AD is a transversal.

In
ECD, we have,

E + C + D = 180o

xo + 45o +
60o = 180o

xo + 105o
= 180o

xo =
180o – 105o = 75o

x = 75

(v)

In
AEF,

Exterior
BED = EAF + EFA

100o = 40o
+ EFA

EFA = 100o – 40o = 60o

Also, CFD = EFA [Vertically
Opposite angles]

CFD = 60o

Now
in FCD,

Exterior
BCF = CFD + CDF

90o = 60o +
xo

xo = 90o
60o = 30o

x = 30

(vi)

In
ABE, we have,

A + B + E = 180o

75o + 65o
+ E = 180o

140o + E = 180o

E = 180o – 140o = 40o

Now, CED = AEB [Vertically
opposite angles]

CED = 40o

Now,
in CED, we have,

C + E + D = 180o

110o + 40o
+ xo = 180o

150o + xo =
180o

xo = 180o
150o = 30o

x = 30

Question 9

In the figure given alongside, AB
CD, EF ∥
BC, ∠BAC
= 60°
and ∠DHF
= 50°.
Find ∠GCH
and ∠AGH.

Solution 9

AB

CD and AC is the transversal.

∠BAC = ∠ACD = 60° (alternate angles)

i.e.
∠BAC
= ∠GCH
= 60°

Now,
∠DHF
= ∠CHG
= 50° (vertically
opposite angles)

In
ΔGCH,
by angle sum property,

GCH + ∠CHG + ∠CGH = 180°

60° + 50° + ∠CGH
= 180°

∠CGH = 70°

Now,
∠CGH
+ ∠AGH
= 180° (linear pair)

70° + ∠AGH = 180°

∠AGH = 110°

Question 10

Calculate the value of x in the given figure.

Solution 10

Produce CD to cut AB at E.

Now, in BDE, we have,

Exterior CDB = CEB + DBE

xo = CEB + 45o     …..(i)

In AEC, we have,

Exterior CEB = CAB + ACE

= 55o + 30o = 85o

Putting CEB = 85o in (i), we get,

xo = 85o + 45o = 130o

x = 130

Question 11

In the given figure, AD divides BAC in the ratio 1: 3 and AD = DB. Determine the value of x.

Solution 11

The angle BAC is divided by AD in the ratio 1 : 3.

Let BAD and DAC be y and 3y, respectively.

As BAE is a straight line,

BAC + CAE = 180o        [linear pair]

BAD + DAC +  CAE = 180o

y + 3y + 108o = 180o

4y = 180o – 108o = 72o

Now, in ABC,

ABC + BCA + BAC = 180o

y + x + 4y = 180o

[Since, ABC = BAD (given AD = DB) and BAC = y + 3y = 4y]

5y + x = 180

5 18 + x = 180

90 + x = 180

x = 180 – 90 = 90

Question 12

The angles of a triangle are in the ratio 2:3:4. Find the angles.

Solution 12

Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o.

Then, 2x + 3x + 4x = 180         [sum of the angles of a triangle is 180o ]

9x = 180

The measures of the required angles are:

2x = (2 20)o = 40o

3x = (3 20)o = 60o

4x = (4 20)o = 80o

Question 13

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so fomed is equal to four right angles.

Solution 13

Given : A ABC in which BC, CA and AB are produced to D, E and F respectively.

To prove : Exterior DCA + Exterior BAE + Exterior FBD = 360o

Proof : Exterior DCA = A + B(i)

Exterior FAE = B + C(ii)

Exterior FBD = A + C(iii)

Adding (i), (ii) and (iii), we get,

Ext. DCA + Ext. FAE + Ext. FBD

= A + B + B + C + A + C

= 2A +2B + 2C

= 2 (A + B + C)

= 2 180o

[Since, in triangle the sum of all three angle is 180o]

= 360o

Hence, proved.

Question 14

In the given figure, show that

A + B + C + D + E + F = 360o.

Solution 14

In ACE, we have,

A + C + E = 180o (i)

In BDF, we have,

B + D + F = 180o (ii)

Adding both sides of (i) and (ii), we get,

A + C+E + B + D + F = 180o + 180o

A + B + C + D + E + F = 360o.

Question 15

In the given figure, AM
BC and AN is the bisector of ∠A.
If ∠ABC
= 70°
and ∠ACB
= 20°,
find ∠MAN.

Solution 15

In
ΔABC, by angle
sum property,

A
+ ∠B
+ ∠C
= 180°

∠A
+ 70° + 20° = 180°

∠A
= 90°

In
ΔABM, by angle
sum property,

BAM
+ ∠ABM
+ ∠AMB
= 180°

∠BAM
+ 70° + 90° = 180°

∠BAM
= 20°

Since AN is the bisector of ∠A,

Now, ∠MAN
+ ∠BAM
= ∠BAN

∠MAN
+ 20° = 45°

∠MAN
= 25°

Question 16

EF, ∠AEF
= 55°
and ∠ACB
= 25°,
find ∠ABC.

Solution 16

∥ EF and EC is the transversal.

∠AEF

= 55°

+ ∠CAB
= 180° (linear pair)

55° + ∠CAB = 180°

∠CAB
= 125°

In ΔABC, by angle
sum property,

ABC
+ ∠CAB
+ ∠ACB
= 180°

∠ABC
+ 125° + 25° = 180°

∠ABC
= 30°

Question 17

In the given figure, ABC is a triangle in which A : B : C = 3 : 2 : 1 and AC CD. Find the measure of

Solution 17

In the given ABC, we have,

A : B : C = 3 : 2 : 1

Let A = 3x, B = 2x, C = x. Then,

A + B + C = 180o

3x + 2x + x = 180o

6x = 180o

x = 30o

A = 3x = 3 30o = 90o

B = 2x = 2 30o = 60o

and, C = x = 30o

Now, in ABC, we have,

Ext ACE = A + B = 90o + 60o = 150o

ACD + ECD = 150o

ECD = 150oACD

ECD = 150o – 90o    [since ]

ECD= 60o

Question 18

In the given figure, AB
DE and BD ∥ FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.

Solution 18

FGH
+ ∠FGE
= 180° (linear pair)

120° + y = 180°

y = 60°

AB
∥ DF and BD is the transversal.

∠ABC
= ∠CDE (alternate angles)

∠CDE
= 50°

BD
∥ FG and DF is the transversal.

∠EFG
= ∠CDE (alternate angles)

∠EFG
= 50°

In ΔEFG, by angle
sum property,

FEG
+ ∠FGE
+ ∠EFG
= 180°

x + y + 50° = 180°

x + 60° + 50° = 180°

x = 70°

Question 19

In the given figure, AB
CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, Find the value of x.

Solution 19

AB
∥ CD and EF is the transversal.

∠AEF
= ∠EFD (alternate angles)

∠AEF
= ∠EFG
+ ∠DFG

65° = ∠EFG + 30°

∠EFG
= 35°

In ΔGEF, by angle
sum property,

GEF
+ ∠EGF
+ ∠EFG
= 180°

x + 90° + 35° = 180°

x = 55°

Question 20

In the given figure, AB
CD, ∠BAE
= 65°
and ∠OEC
= 20°.
Find ∠ECO.

Solution 20

AB
∥ CD and AE is the transversal.

∠BAE
= ∠DOE (corresponding angles)

∠DOE
= 65°

Now,
∠DOE
+ ∠COE
= 180° (linear pair)

65° + ∠COE = 180°

∠COE
= 115°

In ΔOCE, by angle
sum property,

OEC
+ ∠ECO
+ ∠COE
= 180°

20° + ∠ECO + 115° = 180°

∠ECO
= 45°

Question 21

In the given figure, AB
CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB
= 35°
and QP ⊥
EF, find the measure of ∠PQH.

Solution 21

AB
∥ CD and EF is the transversal.

∠EGB
= ∠GHD (corresponding angles)

∠GHD
= 35°

Now,
∠GHD
= ∠QHP (vertically opposite angles)

∠QHP
= 35°

In DQHP, by angle
sum property,

PQH
+ ∠QHP
+ ∠QPH
= 180°

∠PQH
+ 35° + 90° = 180°

∠PQH
= 55°

Question 22

In the given figure, AB
CD and EF ⊥ AB. If EG is the transversal such that ∠GED
= 130°,
find ∠EGF.

Solution 22

AB
∥ CD and GE is the transversal.

∠EGF
+ ∠GED
= 180° (interior
angles are supplementary)

∠EGF
+ 130° = 180°

∠EGF
= 50°

Question 23

In
ABC, if 3A = 4B = 6C, calculate A, B and C.

Solution 23

Let
3A = 4B = 6C = x (say)

Then, 3A = x

A =

4B = x

and 6C = x

C =

As
A + B + C = 180o

A =

B =

C =

Question 24

In
ABC, if A + B = 108o and B + C = 130o, find A, B and C.

Solution 24

A + B = 108o [Given]

But
as A, B and C are the angles of a triangle,

A + B + C = 180o

108o + C = 180o

C = 180o – 108o = 72o

Also, B + C = 130o [Given]

B + 72o = 130o

B = 130o – 72o = 58o

Now
as, A + B = 108o

A + 58o = 108o

A = 108o – 58o = 50o

A = 50o, B = 58o and C = 72o.

Question 25

In
ABC, A + B = 125o and A + C = 113o. Find A, B and C.

Solution 25

Since.
A , B and C are the angles of a triangle .

So, A + B + C = 180o

Now, A + B = 125o [Given]

125o + C = 180o

C = 180o – 125o = 55o

Also, A + C = 113o [Given]

A + 55o = 113o

A = 113o – 55o = 58o

Now
as A + B = 125o

58o + B = 125o

B = 125o – 58o = 67o

A = 58o, B = 67o and C = 55o.

Question 26

In PQR, if P – Q = 42o and Q – R = 21o, find P, Q and R.

Solution 26

Since, P, Q and R are the angles of a triangle.

So,P + Q + R = 180o(i)

Now,P – Q = 42o[Given]

P = 42o + Q(ii)

andQ – R = 21o[Given]

R = Q – 21o(iii)

Substituting the value of P and R from (ii) and (iii) in (i), we get,

42o + Q + Q + Q – 21o = 180o

3Q + 21o = 180o

3Q = 180o – 21o = 159o

Q =

P = 42o + Q

= 42o + 53o = 95o

R = Q – 21o

= 53o – 21o = 32o

P = 95o, Q = 53o and R = 32o.

Question 27

The sum of two angles of a triangle is 116o and their difference is 24o. Find the measure of each angle of the triangle.

Solution 27

Given that the sum of the angles A and B of a ABC is 116o, i.e., A + B = 116o.

Since, A + B + C = 180o

So, 116o + C = 180o

C = 180o – 116o = 64o

Also, it is given that:

A – B = 24o

A = 24o + B

Putting, A = 24o + B in A + B = 116o, we get,

24o + B + B = 116o

2B + 24o = 116o

2B = 116o – 24o = 92o

B =

Therefore, A = 24o + 46o = 70o

A = 70o, B = 46o and C = 64o.

Question 28

Two angles of a triangle are equal and the third angle is greater than each one of them by 18o. Find the angles.

Solution 28

Let the two equal angles, A and B, of the triangle be xo each.

We know,

A + B + C = 180o

xo + xo + C = 180o

2xo + C = 180o(i)

Also, it is given that,

C = xo + 18o(ii)

Substituting C from (ii) in (i), we get,

2xo + xo + 18o = 180o

3xo = 180o – 18o = 162o

x =

Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.

Question 29

Of
the three angles of triangle, one is twice the smallest and another one is
thrice the smallest. Find the angles.

Solution 29

Let C be the smallest angle of ABC.

Then,
A = 2C and B = 3C

Also,
A + B + C = 180o

2C + 3C + C = 180o

6C = 180o

C = 30o

So, A = 2C = 2 30o
= 60o

B = 3C = 3 30o
= 90o

The required angles of the triangle are 60o,
90o, 30o.

## Chapter 8 – Triangles Exercise MCQ

Question 1

Solution 1

Question 2

In a ΔABC it is given that ∠A:∠B:∠C = 3:2:1 and ∠ACD = 90o. If is produced to E Then ∠ECD =?

1. 60°
2. 50°
3. 40°
4. 25°

Solution 2

Question 3

In
the given figure , BO and CO are the bisectors of ∠B
and ∠C
respectively. If ∠A
= 50°,
then ∠BOC=
?

1. 130°
2. 100°
3. 115°
4. 120°

Solution 3

Question 4

In the given figure, side BC of ΔABC
has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠ACD = 7y°. Then, the value of x is

(a) 60

(b) 50

(c) 45

(d) 35

Solution 4

Correct
option: (a)

ACB
+ ∠ACD
= 180° (linear pair)

5y + 7y = 180°

12y = 180°

y = 15°

Now, ∠ACD
= ∠ABC
+ ∠BAC (Exterior angle property)

7y = x + 3y

7(15°) = x + 3(15°)

105° = x + 45°

x = 60°

Question 5

In a ΔABC, if ∠A – ∠B = 42° and ∠B – ∠C = 21° then ∠B = ?

(a) 32°

(b) 63°

(c) 53°

(d) 95°

Solution 5

Correct
option: (c)

A – ∠B = 42°

∠A
= ∠B
+ 42°

B
– ∠C
= 21°

∠C
= ∠B
– 21°

In ΔABC,

A
+ ∠B
+ ∠C
= 180°

∠B
+ 42° + ∠B + ∠B – 21° = 180°

3∠B
= 159

∠B
= 53°

Question 6

In a ΔABC, side BC is produced to D. If ∠ABC
= 50°
and ∠ACD
= 110°
then ∠A
= ?

(a) 160°

(b) 60°

(c) 80°

(d) 30°

Solution 6

Correct
option: (b)

ACD
= ∠B
+ ∠A (Exterior angle property)

110° = 50° + ∠A

∠A
= 60°

Question 7

Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?

1. 50°
2. 55°
3. 65°
4. 75°

Solution 7

Correct option: (d)

Question 8

In the given figure, sides CB and BA of ΔABC have been produced to D and E respectively such that ∠ABD = 110° and ∠CAE = 135°. Than ∠ACB =?

1. 65°
2. 45°
3. 55°
4. 35°

Solution 8

Correct option: (a)

Question 9

The sides BC, CA and AB of ΔABC have been produced to D,E and F respectively.  ∠BAE + ∠CBF + ∠ACD =?

1. 240°
2. 300°
3. 320°
4. 360°

Solution 9

Question 10

BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF
= 30°.
Also, in ΔBAC, ∠BAC = x° and ∠ABC = (x + 10. Then, the value
of x is

(a) 20

(b) 25

(c) 30

(d) 35

Solution 10

Correct
option: (b)

EAF

= 30°

In ΔABD, by angle
sum property

A
+ ∠B
+ ∠D
= 180°

(x + 30)° + (x + 10)° + 90° = 180°

2x + 130° = 180°

2x = 50°

x = 25°

Question 11

In the given figure, two rays BD and CE intersect at a
point A. The side BC of
ΔABC have been produced on both sides to points F and G
respectively. If ∠ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?

(a) x
+ y – 180

(b) x
+ y + 180

(c) 180
– (x + y)

(d) x
+ y + 360°

Solution 11

Correct
option: (a)

ABF
+ ∠ABC
= 180° (linear pair)

x + ∠ABC
= 180°

∠ABC
= 180° – x

ACG
+ ∠ACB
= 180° (linear pair)

y + ∠ACB
= 180°

∠ACB
= 180° – y

In ΔABC, by angle
sum property

ABC
+ ∠ACB
+ ∠BAC
= 180°

(180° – x) + (180° – y) + ∠BAC
= 180°

∠BAC
– x – y + 180° = 0

∠BAC
= x + y – 180°

= ∠BAC (vertically opposite angles)

z = x + y – 180°

Question 12

In the given figure, lines AB and CD intersect at a point O. The sides CA and
OB have been produced to E and F respectively. such that
∠OAE
= x°
and ∠
DBF = y°.

If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?

(a) 190°

(b) 230°

(c) 210°

(d) 270°

Solution 12

Correct
option: (b)

In ΔOAC, by angle sum property

OCA + ∠COA + ∠CAO = 180°

80° + 40° + ∠CAO = 180°

⇒∠CAO = 60°

CAO + ∠OAE = 180° (linear pair)

60° + x = 180°

x = 120°

COA = ∠BOD (vertically opposite angles)

∠BOD
= 40°

In ΔOBD, by angle sum property

OBD + ∠BOD + ∠ODB = 180°

∠OBD + 40° + 70° = 180°

∠OBD
= 70°

OBD + ∠DBF = 180° (linear pair)

70° + y = 180°

y = 110°

x + y = 120° + 110° = 230°

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