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Page 1
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
Page 2
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
Page 3
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75°
Q u e s t i o n : 3 4
In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ?
a
65°
b
45°
c
55°
d
35°
S o l u t i o n :
a 65°
We have :
?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70°
Side AB of triangle ABC is produced to E.
? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65°
Q u e s t i o n : 3 5
The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ?
a 240°
b 300°
c 320°
d 360°
S o l u t i o n :
d 360°
We have :
?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii)
Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540°
? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360°
Q u e s t i o n : 3 6
The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is
a 20
b 25
c 30
d 35
S o l u t i o n :
In the given figure, ?CAD = ?EAF Verticallyoppositeangles
? ?CAD = 30°
In ?ABD,
?ABD + ?BAD + ?ADB = 180° Anglesumproperty
? (x + 10)° + (x° + 30°) + 90° = 180°
? 2x° + 130° = 180°
? 2x° = 180° - 130° = 50°
? x = 25
Thus, the value of x is 25.
Page 4
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75°
Q u e s t i o n : 3 4
In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ?
a
65°
b
45°
c
55°
d
35°
S o l u t i o n :
a 65°
We have :
?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70°
Side AB of triangle ABC is produced to E.
? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65°
Q u e s t i o n : 3 5
The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ?
a 240°
b 300°
c 320°
d 360°
S o l u t i o n :
d 360°
We have :
?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii)
Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540°
? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360°
Q u e s t i o n : 3 6
The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is
a 20
b 25
c 30
d 35
S o l u t i o n :
In the given figure, ?CAD = ?EAF Verticallyoppositeangles
? ?CAD = 30°
In ?ABD,
?ABD + ?BAD + ?ADB = 180° Anglesumproperty
? (x + 10)° + (x° + 30°) + 90° = 180°
? 2x° + 130° = 180°
? 2x° = 180° - 130° = 50°
? x = 25
Thus, the value of x is 25.
Hence, the correct answer is option b
.
Q u e s t i o n : 3 7
In the given figure, two rays BD and CE intersect at a point A. The side BC of ?ABC 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°
S o l u t i o n :
In the given figure, ?ABF + ?ABC = 180° Linearpairofangles
? x° + ?ABC = 180°
? ?ABC = 180° - x° .....1
Also, ?ACG + ?ACB = 180° Linearpairofangles
? y° + ?ACB = 180°
? ?ACB = 180° - y° .....2
Also, ?BAC = ?DAE = z° .....3
Verticallyoppositeangles
In ?ABC,
?BAC + ?ABC + ?ACB = 180° Anglesumproperty
? z° + 180° - x° + 180° - y° = 180°
Using(1), (2)and(3)
? z = x + y - 180
Hence, the correct answer is option a
.
Q u e s t i o n : 3 8
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°
S o l u t i o n :
In the given figure, ?BOD = ?COA Verticallyoppositeangles
? ?BOD = 40° .....1
In ?ACO,
?OAE = ?OCA + ?COA Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles
? x° = 80° + 40° = 120° .....2
In ?BDO,
Page 5
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75°
Q u e s t i o n : 3 4
In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ?
a
65°
b
45°
c
55°
d
35°
S o l u t i o n :
a 65°
We have :
?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70°
Side AB of triangle ABC is produced to E.
? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65°
Q u e s t i o n : 3 5
The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ?
a 240°
b 300°
c 320°
d 360°
S o l u t i o n :
d 360°
We have :
?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii)
Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540°
? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360°
Q u e s t i o n : 3 6
The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is
a 20
b 25
c 30
d 35
S o l u t i o n :
In the given figure, ?CAD = ?EAF Verticallyoppositeangles
? ?CAD = 30°
In ?ABD,
?ABD + ?BAD + ?ADB = 180° Anglesumproperty
? (x + 10)° + (x° + 30°) + 90° = 180°
? 2x° + 130° = 180°
? 2x° = 180° - 130° = 50°
? x = 25
Thus, the value of x is 25.
Hence, the correct answer is option b
.
Q u e s t i o n : 3 7
In the given figure, two rays BD and CE intersect at a point A. The side BC of ?ABC 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°
S o l u t i o n :
In the given figure, ?ABF + ?ABC = 180° Linearpairofangles
? x° + ?ABC = 180°
? ?ABC = 180° - x° .....1
Also, ?ACG + ?ACB = 180° Linearpairofangles
? y° + ?ACB = 180°
? ?ACB = 180° - y° .....2
Also, ?BAC = ?DAE = z° .....3
Verticallyoppositeangles
In ?ABC,
?BAC + ?ABC + ?ACB = 180° Anglesumproperty
? z° + 180° - x° + 180° - y° = 180°
Using(1), (2)and(3)
? z = x + y - 180
Hence, the correct answer is option a
.
Q u e s t i o n : 3 8
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°
S o l u t i o n :
In the given figure, ?BOD = ?COA Verticallyoppositeangles
? ?BOD = 40° .....1
In ?ACO,
?OAE = ?OCA + ?COA Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles
? x° = 80° + 40° = 120° .....2
In ?BDO,
?DBF = ?BDO + ?BOD Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles
? y° = 70° + 40° = 110°
Using(1)
.....3
Adding 2
and 3
, we get
x° + y° = 120° + 110° = 230°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 9
In a ?ABC, it is given that ?A : ?B : ?C = 3 : 2 : 1 and ?ACD = 90°. If BC is produced to E then ?ECD = ?
a 60°
b 50°
c 40°
d 25°
S o l u t i o n :
Let ?A = (3x)°, ?B = (2x)° and ?C = x°
Then,
3x +2x +x = 180° [Sum of the angles of a triangle] ? 6x = 180° ? x = 30°
Hence, the angles are
?A = 3 ×30° = 90°, ?B = 2 ×30° = 60° and ?C = 30°
Side BC of triangle ABC is produced to E.
? ?ACE = ?A + ?B ? ?ACD + ?ECD = 90°+60° ? 90°+ ?ECD = 150° ? ?ECD = 60°
Hence, the correct answer is option a
.
Q u e s t i o n : 4 0
In the given figure, BO and CO are the bisectors of ?B and ?C respectively. If ?A = 50° then ?BOC = ?
a 130°
b 100°
c 115°
d 120°
S o l u t i o n :
c 115°
In ? ABC
, we have:
?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? 50°+ ?B + ?C = 180° ? ?B + ?C = 130° ?
1
2
?B +
1
2
?C = 65° . . . (i)
In ? OBC
, we have:
?OBC + ?OCB + ?BOC = 180° ?
1
2
?B +
1
2
?C + ?BOC = 180° [Using (i)] ? 65°+ ?BOC = 180° ? ?BOC = 115°
Q u e s t i o n : 4 1
In the given figure, side BC of ?ABC has been produced to a point D. If ?A = 3y°, ?B = x°, ?C = 5y° and ?CBD = 7y°. Then, the value of x is
a 60
b 50
c 45
d 35
S o l u t i o n :
D i s c l a i m e r: In the question ?ACD should be 7y°.
In the given figure, ?ACB + ?ACD = 180° Linearpairofangles
? 5y° + 7y° = 180°
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RS Aggarwal MCQs: Triangles Notes
RS Aggarwal MCQs: Triangles Notes offer in-depth insights into the specific topic to help you master it with ease.
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RS Aggarwal MCQs: Triangles Class 9 Questions
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The content of RS Aggarwal MCQs: Triangles is prepared as per the latest Class 9 syllabus.
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