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

## Chapter 8 – Triangles Exercise Ex. 8

In _{}ABC, if _{}B = 76^{o} and _{}C = 48^{o}, find _{}A.

Since,

sum of the angles of a triangle is 180^{o}

_{}A + _{}B + _{}C = 180^{o}

_{} _{}A + 76^{o}

+ 48^{o} = 180^{o}

_{} _{}A = 180^{o} – 124^{o} = 56^{o}

_{} _{}A = 56^{o}

In a right-angled triangle, one of the acute angles measures 53^{o}.

Find the measure of each angle of the triangle.

Let ABC be a right angled triangle and _{}C = 90^{o}

Since,

_{}A + _{}B + _{}C = 180^{o}

_{} _{}A + _{}B = 180^{o} – _{}C = 180^{o} – 90^{o} = 90^{o}

Suppose _{}A = 53^{o}

Then, 53^{o} + _{}B = 90^{o}

_{} _{}B = 90^{o} – 53^{o} = 37^{o}

_{} The required angles are 53^{o}, 37^{o}

and 90^{o}.

In

a right-angled triangle, one of the acute angles measures 53^{o}.

Find the measure of each angle of the triangle.

Let

ABC be a right angled triangle and _{}C = 90^{o}

Since,

_{}A + _{}B + _{}C = 180^{o}

_{} _{}A + _{}B = 180^{o} – _{}C = 180^{o} – 90^{o} = 90^{o}

Suppose _{}A = 53^{o}

Then, 53^{o} + _{}B = 90^{o}

_{} _{}B = 90^{o} – 53^{o} = 37^{o}

_{} The required angles are 53^{o}, 37^{o}

and 90^{o}.

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

Let ABC be a triangle.

So, A < B + C

Adding A to both sides of the inequality,

2 A < A + B + C

2 A < 180^{o} [Since A + B + C = 180^{o}]

Similarly, B <A + C

B < 90^{o}

and C < A + B

C < 90^{o}

ABC is an acute angled triangle.

If

one angle of a triangle is greater than the sum of the other two, show that

the triangle is obtuse angled.

Let

ABC be a triangle and _{}B > _{}A + _{}C

Since,

_{}A + _{}B + _{}C = 180^{o}

_{} _{}A + _{}C = 180^{o} – _{}B

_{Therefore, we get,}

_{}B > 180^{o }– _{}B

Adding

_{}B on both sides of the inequality, we get,

_{}B + _{}B > 180^{o} – _{}B + _{}B

_{} 2_{}B > 180^{o}

_{} _{}B > _{}

i.e.,

_{}B > 90^{o} which means _{}B is an obtuse angle.

_{} _{}ABC is an obtuse angled triangle.

In

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

Since

_{}ACB and _{}ACD form a linear pair.

So, _{}ACB + _{}ACD = 180^{o}

_{} _{}ACB + 128^{o} = 180^{o}

_{} _{}ACB = 180^{o} – 128 = 52^{o}

Also, _{}ABC + _{}ACB + _{}BAC = 180^{o}

_{} 43^{o} + 52^{o}

+ _{}BAC = 180^{o}

_{} 95^{o} + _{}BAC = 180^{o}

_{} _{}BAC = 180^{o} – 95^{o} = 85^{o}

_{} _{}ACB = 52^{o} and _{}BAC = 85^{o}.

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 = 106^{o} and _{}ACE= 118^{o}, find the measure of each angle of the triangle.

As _{}DBA and _{}ABC form a linear pair.

So,_{}DBA + _{}ABC = 180^{o}

_{}106^{o} + _{}ABC = 180^{o}

_{}_{}ABC = 180^{o} – 106^{o} = 74^{o}

Also, _{}ACB and _{}ACE form a linear pair.

So,_{}ACB + _{}ACE = 180^{o}

_{}_{}ACB + 118^{o} = 180^{o}

_{}_{}ACB = 180^{o} – 118^{o} = 62^{o}

In _{}ABC, we have,

_{}ABC + _{}ACB + _{}BAC = 180^{o}

74^{o} + 62^{o} + _{}BAC = 180^{o}

_{}136^{o} + _{}BAC = 180^{o}

_{}_{}BAC = 180^{o} – 136^{o} = 44^{o}

_{}In triangle ABC, _{}A = 44^{o}, _{}B = 74^{o} and _{}C = 62^{o}

Calculate

the value of x in each of the following figures.

(i)

_{}

(ii) _{}

(iii)

_{}

Given:

AB || CD

(vi)

_{}

(i) _{}EAB + _{}BAC = 180^{o} [Linear pair angles]

110^{o} + _{}BAC = 180^{o}

_{} _{}BAC = 180^{o} – 110^{o} = 70^{o}

Again, _{}BCA + _{}ACD = 180^{o }[Linear

pair angles]

_{} _{}BCA + 120^{o}

= 180^{o}

_{} _{}BCA = 180^{o} – 120^{o} = 60^{o}

Now,

in _{}ABC,

_{}ABC + _{}BAC + _{}ACB = 180^{o}

x^{o} + 70^{o}

+ 60^{o} = 180^{o}

_{} x + 130^{o}

= 180^{o}

_{} x = 180^{o} – 130^{o}

= 50^{o}

_{} x = 50

(ii)

In

_{}ABC,

_{ }A

+ _{}B + _{}C = 180^{o}

_{} 30^{o} + 40^{o}

+ _{}C = 180^{o}

_{} 70^{o}

+ _{}C = 180^{o}

_{} _{}C = 180^{o} – 70^{o} = 110^{o}

Now _{}BCA + _{}ACD = 180^{o} [Linear pair]

_{} 110^{o}

+ _{}ACD = 180^{o}

_{}

_{}ACD = 180^{o} – 110^{o} = 70^{o}

In

_{}ECD,

_{}ECD + _{}CDE + _{}CED = 180^{o}

_{} 70^{o} + 50^{o}

+ _{}CED = 180^{o}

_{} 120^{o}

+ _{}CED = 180^{o}

_{}

_{}CED = 180^{o} – 120^{o} = 60^{o}

Since

_{}AED and _{}CED from a linear pair

So, _{}AED + _{}CED = 180^{o}

_{} x^{o}

+ 60^{o} = 180^{o}

_{} x^{o}

= 180^{o} – 60^{o} = 120^{o}

_{} x =

120

(iii)

_{ }EAF

= _{}BAC

[Vertically opposite angles]

_{} _{}BAC = 60^{o}

In

_{}ABC, exterior _{}ACD is equal to the sum of two opposite interior

angles.

So, _{}ACD = _{}BAC + _{}ABC

_{} 115^{o} = 60^{o}

+ x^{o}

_{} x^{o} = 115^{o}

– 60^{o} = 55^{o}

_{} x = 55

(iv)

Since

AB || CD and AD is a transversal.

So, _{}BAD = _{}ADC

_{} _{}ADC = 60^{o}

In

_{}ECD, we have,

_{}E + _{}C + _{}D = 180^{o}

_{} x^{o} + 45^{o} +

60^{o} = 180^{o}

_{} x^{o} + 105^{o}

= 180^{o}

_{} x^{o} =

180^{o} – 105^{o} = 75^{o}

_{} x = 75

(v)

In

_{}AEF,

Exterior

_{}BED = _{}EAF + _{}EFA

_{} 100^{o} = 40^{o}

+ _{}EFA

_{} _{}EFA = 100^{o} – 40^{o} = 60^{o}

Also, _{}CFD = _{}EFA [Vertically

Opposite angles]

_{} _{}CFD = 60^{o}

Now

in _{}FCD,

Exterior

_{}BCF = _{}CFD + _{}CDF

_{} 90^{o} = 60^{o} +

x^{o}

_{} x^{o} = 90^{o} –

60^{o} = 30^{o}

_{} x = 30

(vi)

In

_{}ABE, we have,

_{}A + _{}B + _{}E = 180^{o}

_{} 75^{o} + 65^{o}

+ _{}E = 180^{o}

_{} 140^{o} + _{}E = 180^{o}

_{} _{}E = 180^{o} – 140^{o} = 40^{o}

Now, _{}CED = _{}AEB [Vertically

opposite angles]

_{} _{}CED = 40^{o}

Now,

in _{}CED, we have,

_{}C + _{}E + _{}D = 180^{o}

_{} 110^{o} + 40^{o}

+ x^{o} = 180^{o}

_{} 150^{o} + x^{o} =

180^{o}

_{} x^{o} = 180^{o} –

150^{o }= 30^{o}

_{} x = 30

In the figure given alongside, AB ∥

CD, EF ∥

BC, ∠BAC

= 60°

and ∠DHF

= 50°.

Find ∠GCH

and ∠AGH.

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°

Calculate the value of x in the given figure.

Produce CD to cut AB at E.

Now, in _{}BDE, we have,

Exterior _{}CDB = _{}CEB + _{}DBE

_{} x^{o} = _{}CEB + 45^{o } …..(i)

In _{} AEC, we have,

Exterior _{}CEB = _{}CAB + _{}ACE

= 55^{o} + 30^{o} = 85^{o}

Putting _{}CEB = 85^{o} in (i), we get,

x^{o} = 85^{o} + 45^{o} = 130^{o}

_{} x = 130

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

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 = 180^{o } [linear pair]

_{} _{}BAD + _{}DAC + _{ }CAE = 180^{o}

_{} y + 3y + 108^{o} = 180^{o}

_{} 4y = 180^{o} – 108^{o} = 72^{o}

_{} _{}

Now, in _{}ABC,

_{}ABC + _{}BCA + _{}BAC = 180^{o}

y + x + 4y = 180^{o}

[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

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

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 180^{o} ]^{ }

_{} 9x = 180

_{} _{}

_{} The measures of the required angles are:

2x = (2 _{} 20)^{o} = 40^{o}

3x = (3 _{} 20)^{o} = 60^{o}

4x = (4 _{} 20)^{o} = 80^{o}

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.

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 = 360^{o}

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

= 2_{}A +2_{}B + 2_{}C

= 2 (_{}A + _{}B + _{}C)

= 2 _{}180^{o}

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

= 360^{o}

Hence, proved.

In the given figure, show that

_{}A + _{}B + _{}C + _{}D + _{}E + _{}F = 360^{o}.

In _{}ACE, we have,

_{}A + _{}C + _{}E = 180^{o} (i)

In _{}BDF, we have,

_{}B + _{}D + _{}F = 180^{o} (ii)

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

_{}A + _{}C+_{}E + _{}B + _{}D + _{}F = 180^{o} + 180^{o}

_{}_{}A + _{}B + _{}C + _{}D + _{}E + _{}F = 360^{o}.

In the given figure, AM ⊥

BC and AN is the bisector of ∠A.

If ∠ABC

= 70°

and ∠ACB

= 20°,

find ∠MAN.

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°

In the given figure, BAD ∥

EF, ∠AEF

= 55°

and ∠ACB

= 25°,

find ∠ABC.

BAD

∥ EF and EC is the transversal.

⇒ ∠AEF

= ∠CAD (corresponding angles)

⇒ ∠CAD

= 55°

Now, ∠CAD

+ ∠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°

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

In the given _{}ABC, we have,

_{}A : _{}B : _{}C = 3 : 2 : 1

Let _{}A = 3x, _{}B = 2x, _{}C = x. Then,

_{}A + _{}B + _{}C = 180^{o}

_{} 3x + 2x + x = 180^{o}

_{ } 6x = 180^{o}

_{} x = 30^{o}

_{} _{}A = 3x = 3 _{} 30^{o} = 90^{o}

_{}B = 2x = 2 _{} 30^{o} = 60^{o}

and, _{}C = x = 30^{o}

Now, in ABC, we have,

Ext _{}ACE = _{}A + _{}B = 90^{o} + 60^{o} = 150^{o}

_{} _{}ACD + _{}ECD = 150^{o}

_{}ECD = 150^{o} – _{}ACD

ECD = 150^{o} – 90^{o} [since , ]

ECD= 60^{o}

In the given figure, AB ∥

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

∠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°

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.

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°

In the given figure, AB ∥

CD, ∠BAE

= 65°

and ∠OEC

= 20°.

Find ∠ECO.

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°

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.

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°

In the given figure, AB ∥

CD and EF ⊥ AB. If EG is the transversal such that ∠GED

= 130°,

find ∠EGF.

AB

∥ CD and GE is the transversal.

⇒ ∠EGF

+ ∠GED

= 180° (interior

angles are supplementary)

⇒ ∠EGF

+ 130° = 180°

⇒ ∠EGF

= 50°

In

_{}ABC, if 3_{}A = 4_{}B = 6_{}C, calculate _{}A, _{}B and _{}C.

Let

3_{}A = 4_{}B = 6_{}C = x (say)

Then, 3_{}A = x

_{} _{}A = _{}

4_{}B = x

_{} _{}

and 6_{}C = x

_{} _{}C = _{}

As

_{}A + _{}B + _{}C = 180^{o}

_{}

_{} _{}A = _{}

_{}B = _{}

_{}C = _{}

In

_{}ABC, if _{}A + _{}B = 108^{o} and _{}B + _{}C = 130^{o}, find _{}A, _{}B and _{}C.

_{}A + _{}B = 108^{o} [Given]

But

as _{}A, _{}B and _{}C are the angles of a triangle,

_{}A + _{}B + _{}C = 180^{o}

_{} 108^{o} + _{}C = 180^{o}

_{} _{}C = 180^{o} – 108^{o} = 72^{o}

Also, _{}B + _{}C = 130^{o} [Given]

_{} _{}B + 72^{o} = 130^{o}

_{} _{}B = 130^{o} – 72^{o} = 58^{o}

Now

as, _{}A + _{}B = 108^{o}

_{} _{}A + 58^{o} = 108^{o}

_{} _{}A = 108^{o} – 58^{o} = 50^{o}

_{} _{}A = 50^{o}, _{}B = 58^{o} and _{}C = 72^{o}.

In

_{}ABC, _{}A + _{}B = 125^{o} and _{}A + _{}C = 113^{o}. Find _{}A, _{}B and _{}C.

Since.

_{}A , _{}B and _{}C are the angles of a triangle .

So, _{}A + _{}B + _{}C = 180^{o}

Now, _{}A + _{}B = 125^{o} [Given]

_{} 125^{o} + _{}C = 180^{o}

_{} _{}C = 180^{o} – 125^{o} = 55^{o}

Also, _{}A + _{}C = 113^{o} [Given]

_{} _{}A + 55^{o} = 113^{o}

_{} _{}A = 113^{o} – 55^{o} = 58^{o}

Now

as _{}A + _{}B = 125^{o}

_{} 58^{o} + _{}B = 125^{o}

_{} _{}B = 125^{o} – 58^{o} = 67^{o}

_{} _{}A = 58^{o}, _{}B = 67^{o} and _{}C = 55^{o}.

In _{}PQR, if _{}P – _{}Q = 42^{o} and _{}Q – _{}R = 21^{o}, find _{}P, _{}Q and _{}R.

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

So,_{}P + _{}Q + _{}R = 180^{o}(i)

Now,_{}P – _{}Q = 42^{o}[Given]

_{}_{}P = 42^{o} + _{}Q(ii)

and_{}Q – _{}R = 21^{o}[Given]

_{}R = _{}Q – 21^{o}(iii)

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

42^{o} + _{}Q + _{}Q + _{}Q – 21^{o} = 180^{o}

_{}3_{}Q + 21^{o} = 180^{o}

_{}3_{}Q = 180^{o} – 21^{o} = 159^{o}

_{}_{}Q = _{}

_{}_{}P = 42^{o} + _{}Q

= 42^{o} + 53^{o} = 95^{o}

_{}R = _{}Q – 21^{o}

= 53^{o} – 21^{o} = 32^{o}

_{}_{}P = 95^{o}, _{}Q = 53^{o} and _{}R = 32^{o}.

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

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

Since, _{}A + _{}B + _{}C = 180^{o}

So, 116^{o} + _{}C = 180^{o}

_{} _{}C = 180^{o} – 116^{o} = 64^{o}

Also, it is given that:

_{}A – _{}B = 24^{o}

_{} _{}A = 24^{o} + _{}B

Putting, _{}A = 24^{o} + _{}B in _{}A + _{}B = 116^{o}, we get,

24^{o} + _{}B + _{}B = 116^{o}

_{} 2_{}B + 24^{o} = 116^{o}

_{} 2_{}B = 116^{o} – 24^{o} = 92^{o}

_{} _{}B = _{}

Therefore, _{}A = 24^{o} + 46^{o} = 70^{o}

_{} _{}A = 70^{o}, _{}B = 46^{o} and _{}C = 64^{o}.

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

Let the two equal angles, _{}A and _{}B, of the triangle be x^{o} each.

We know,

_{}A + _{}B + _{}C = 180^{o}

_{}x^{o} + x^{o} + _{}C = 180^{o}

_{}2x^{o} + _{}C = 180^{o}(i)

Also, it is given that,

_{}C = x^{o} + 18^{o}(ii)

Substituting _{}C from (ii) in (i), we get,

2x^{o} + x^{o} + 18^{o} = 180^{o}

_{}3x^{o} = 180^{o} – 18^{o} = 162^{o}

_{}x = _{}

Thus, the required angles of the triangle are 54^{o}, 54^{o} and x^{o} + 18^{o} = 54^{o} + 18^{o} = 72^{o}.

Of

the three angles of triangle, one is twice the smallest and another one is

thrice the smallest. Find the angles.

Let _{}C be the smallest angle of _{}ABC.

Then,

_{}A = 2_{}C and _{}B = 3_{}C

Also,

_{}A + _{}B + _{}C = 180^{o}

_{} 2_{}C + 3_{}C + _{}C = 180^{o}

_{} 6_{}C = 180^{o}

_{} _{}C = 30^{o}

So, _{}A = 2_{}C = 2 _{} 30^{o}

= 60^{o}

_{}B = 3_{}C = 3 _{} 30^{o}

= 90^{o}

_{} The required angles of the triangle are 60^{o},

90^{o}, 30^{o}.

## Chapter 8 – Triangles Exercise MCQ

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

- 60°
- 50°
- 40°
- 25°

In

the given figure , BO and CO are the bisectors of ∠B

and ∠C

respectively. If ∠A

= 50°,

then ∠BOC=

?

- 130°
- 100°
- 115°
- 120°

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

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°

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

(a) 32°

(b) 63°

(c) 53°

(d) 95°

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°

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°

Correct

option: (b)

∠ACD

= ∠B

+ ∠A (Exterior angle property)

⇒

110° = 50° + ∠A

⇒ ∠A

= 60°

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 = ?

- 50°
- 55°
- 65°
- 75°

Correct option: (d)

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 =?

- 65°
- 45°
- 55°
- 35°

Correct option: (a)

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

- 240°
- 300°
- 320°
- 360°

In the given figure, EAD ⊥

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

Correct

option: (b)

∠EAF

= ∠CAD (vertically opposite angles)

⇒ ∠CAD

= 30°

In ΔABD, by angle

sum property

∠A

+ ∠B

+ ∠D

= 180°

⇒

(x + 30)° + (x + 10)° + 90° = 180°

⇒

2x + 130° = 180°

⇒

2x = 50°

⇒

x = 25°

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°

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°

Now, ∠EAD

= ∠BAC (vertically opposite angles)

⇒

z = x + y – 180°

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°

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°