Download LINES AND ANGLES

CHAPTER
6
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LINES AND ANGLES
Points to Remember :
1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. This
property is known as the Linear Pair Axiom.
2. If two lines intersect each other, then the vertically opposite angles are equal.
3. If a transversal intersects two parallel lines, then
(a) each pair of corresponding angles is equal.
(b) each pair of alternate interior angles is equal.
(c) each pair of interior angles on the same side of the transversal is supplementary.
4. If a transversal intersects two line such that, either
(a) any one pair of corresponding angles is equal, or
(b) any one pair of alternate interior angles is equal, or
(c) any one pair of interior angles on the same side of of the transversal is supplementary, then the lines
are parallel.
5. Two intersecting lines cannot both be parallel to the same line.
6. Lines which are parallel to a given line are parallel to each other.
7. The sum of three angles of a triangle is 180°.
8. If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior
opposite angles.
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ILLUSTRATIVE EXAMPLES
Example 1. If two lines intersects, prove that vertically opposite angles are equal.
Solution.
Given : Two lines AB and CD intersects at O.
To prove : 1 = 3 and 2 = 4
Proof : OB stands on line COD.
1 + 2 = 180°
( linear pair) ...(1)
Also, OD stands on line AOB
2 + 3 = 180°
( linear pair) ...(2)
from (1) and (2)
1 + 2 = 2 + 3
1 = 3
Similarly, 2 = 4
Hence proved
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Example 2. The complement of an angle is half of itself. Find the angle and its complement.
Solution.
Let the given angle be x. Then, its complement is 90° – x.
According to given question,
90° – x =
1
x
2

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x
x
 90
2
3x
2  90
 90  x 
2
3
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55

x = 60°
Hence, the given angle is 60° and its complement is 90° – 60° i.e. 30°.
Example 3. Two supplementary angles are in the ratio 3 : 2. Find the angles.
Solution.
Let the angles be 3x and 2x.
according to given question,
3x + 2x = 180°
( angles are supplementary)
 x
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180º
 36º
5

5x = 180°

Angles are 3 x  3  36º  108  
 Ans.
and 2 x  2  36º  72  
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Example 4. What value of x would make AOB a line in given figure if AOC = (3x –10)° and
BOC = (7x + 30)°?
Solution.




B
Since, OB and OA are opposite rays
BOC + AOC = 180º
7x + 30° + 3x – 10° = 180°
10x + 20° = 180° 10x = 180° – 20°
10x = 160º  x 
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160
10
 x  16 Ans.
Example 5. Rays OA, OB, OC, OD and OE have the common initial point O. Show that :
AOB + BOC + COD + DOE + EOA = 360°
Solution.
Let us draw a ray OF oposite to ray OA.
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B
C
3
4
F
D
2
0
5
1
6
A
E
so, 1 + 2 + 3 = 180°
and
4 + 5 + 6 = 180°
( linear pair axiom)
adding (1) and (2),
(1 + 2 + 3) + (4 + 5 + 6) = 180° + 180°
 1 + 2 + (3+ 4) + 5 + 6 = 360°
 AOB + BOC + COD + DOE + EOA = 360°
Hence shown.
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LINES AND ANGLES
...(1)
...(2)
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Example 6. Prove that if a transversal intersects two parallel lines, then each pair of interior angle on the same
side of the transversal is supplementary.
Solution.
Given : A transversal l intersects two parallel lines AB and CD at P and Q.
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4 + 5 = 180° and
3 + 6 = 180º
Proof : Ray QD stands on line l

1 + 4 = 180°
...(1)
( linear pair axiom)
also,
1 = 5
...(2)
( corresponding angles)
from (1) and (2)
4 + 5 = 180°
Again, ray QC stands on line l

2 + 3 = 180°
...(3)
( linear pair axiom)
and,
2 = 6
...(4)
( corresponding angles)
from (3) and (4)
3 + 6 = 180° Hence proved.
Example 7. If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of
interior angles encloses a rectangle.
Solution.
Given : Two parallel lines AB and CD are intersected by a transversal  at P and R respectively.
PQ, RQ, RS and PS are bisectors of APR, PRC, PRD and BPR respectively.
To prove :
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B
To prove : PQRS is a rectangle
Proof : Since AB || CD and l is a transversal
 APR = PRD ( alt. interior angles)

1
1
APR = PRD  QPR = PRS
2
2
But, these are alternate interior angles.
 PQ || RS. Similarly QR || PS.
 PQRS is a parallelogram.
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Now, ray PR stands on AB
 APR + BPR = 180°
( linear pair)
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1
1
APR + BPR = 90°
2
2
 QPR + SPR = 90°
 QPS = 90°
Thus, PQRS is a parallelogram, one of whose angle is 90°
 PQRS is a rectangle.
Hence proved.
Example 8. In the given figure lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.

B
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—NCERT
Since XY is a line  a + b + 90° = 180° ( linear pair)
 a + b = 180° – 90°

a + b = 90°
But a : b = 2 : 3. Let a = 2x and b = 3x.
 2x + 3x = 90°  5x = 90°  x = 18°
 a = 2x = 2 × 18° = 36° and b = 3x = 3 × 18° = 54°
Now, OM and ON are opposite rays.  MON is a line.
Since ray OX stands on MN  MOX + XON = 180° ( linear pair)

c + b = 180°
 c + 54° = 180°  C = 180° – 154° = 126°

c = 126° Ans.
Example 9. It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information.
If ray YQ bisects ZYP, find XYQ and reflex QYP.
—NCERT
Solution.
Since XY is produced to point P,  XP is a straight line.
Since, YZ stands on XP,  XYZ + ZYP = 180° ( linear pair)
 64° + ZYP =1 80°
 ZYP = 180° – 64° = 116°
Since ray YQ bisects ZYP
Solution.
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116
 58
2
Now, XYQ = XYZ + ZYQ
 XYQ = 64° + 58° = 122°
and reflex QYP = 360° – QYP = 360° – 58° = 302° Ans.

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QYP  ZYQ 
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Example 10. In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
—NCERT
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Since, AB || CD and CD || EF  AB || CD || EF.
Now, CD || EF and PR is a transversal.
 PQD = PRF
 180° – y = z

y + z = 180°
also, y : z = 3 : 7. Let y = 3a and z = 7a
 3a + 7a = 180°  10a = 180°  a = 18°

y = 3 × 18° = 54° and z = 7 × 18° = 126°
Now, AB || CD and PQ is a transversal.

x + y = 180°
( consecutive interior angles are supplementary)
 x + 54° = 180°

x = 180° – 54° = 126°

x = 126° Ans.
Example 11. In the given figure, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY
respectively of XYZ, find OZY and YOZ.
—NCERT
Solution.
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Solution.
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In XYZ, YXZ + XYZ + XZY = 180°
( angle sum property of a triangle)
 62° + 54° + XZY = 180°
 XZY = 180° – 62° – 54° = 64°
Since, YO and ZO are bisectors of XYZ and XZY,

OYZ 
1
1
XYZ   54  27
2
2
1
1
XZY   64  32
2
2
In OYZ, we have YOZ + OYZ + OZY = 180° ( angle sum property of a triangle)
 YOZ + 27° + 32° = 180°
 YOZ = 180° – 27° – 32° = 121°
Hence, OZY = 32° and YOZ = 121° Ans.
and, OZY 
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Example 12. In the given figure, if PQ  PS, PQ || SR, SQR = 28° and QRT = 65°, then find the values of x
and y.
—NCERT
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QRT = RQS + QSR
( Exterior angle property in SRQ)
 65° = 28° + QSR
 QSR = 65° – 28° = 37°
Since, PQ || SR and the transversal PS intersects then at P and S respectively.
 PSR + SPQ = 180°
( sum of consecutive interior angles is 180°)
 (PSQ + QSR) + 90° = 180°
 y + 37° + 90° = 180°
 y = 180° – 37° – 90° = 53°
Now, in the right SPQ, we have PQS + PSQ = 90°
 x + 53°= 90°
 x = 90° – 53° = 37°
Hence, x = 37° and y = 53° Ans.
Example 13. In each of the following figures, AB || CD, find x.
Solution.
B
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B
50°
x°
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C
(i)
Solution.
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25°
D
(i) Through O, Draw EOF || AB || CD.
then, 1 + 2 = x
Now, EO || AB and BO is the transversal.
 1 + 50º = 180º
( Interior angles on the same side of the transversal)
A
B
50°
E
1
O
2
C
F
25°
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MATHEMATICS–IX
 1 = 180° – 50° 1 = 130°
Again, EO || CD and OD is a transversal
 2 + 25° = 180°
 2 = 180° – 25° 2 = 155°
adding (1) and (2)
1 + 2 = 130° + 155° = 285°
...(1)
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...(2)
i.e. x  285º Ans.
(ii) Produce AB to intersect CE and F.
A
105°
B
125°
G
F
x°
E
Now, ABE + EBF = 180°
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C
( Linear pair)
 125° + EBF = 180°
 EBF = 180° – 125° = 55°
Again, CD || FG and CF is a transversal
DCF + CFG = 180° ( Interior angles on same side of transversal)
105° + CFG = 180º
CFG = 180° – 105° = 75°
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Also, BFE = CFG = 75°
B
( vertically opposite angles)
Now, In BEF,
BEF + EBF + BFE = 180° ( angle sum property)
 x° + 55° + 75° = 180°
 xº + 130° = 180°
 x = 180º – 130°
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
x  50 Ans.
Example 14. The side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point
1
T, then prove that QTR = QPR.
—NCERT
2
Solution.
Side QR of PQR is produced to S,
 Ext. PRS =P + Q
( Exterior angle sum property of a triangle)

1
1
1
Ext PRS = P + Q
2
2
2
1
P + 1
...(1)
2
Again, In QRT.,
Ext. TRS = T + 1 ( same as above)
 2 = T + 1
...(2)

T
P
2 =
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2
1
1
Q
LINES AND ANGLES
2
R
S
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Equating (1) and (2), we get
1
P + 1 = T + 1
2
1
1
 T = P or QTR = QPR
2
2
Hence Proved.
Example 15. Bisectors of exterior angles of ABC (obtained by producing sides AB and AC) meet at O. Prove
1
that BOC = 90° – A.
2
Solution.
Given : ABC in which bisectors of exterior angles meet at O (as shown)
1
To prove : BOC = 90 – A
2
Proof : In OBC
1 + 2 + 0 = 180º ( angle sum property of a triangle)
1
1

CBD + BCE + O = 180°
2
2
 CBD + BCE + 2O = 360°
 (180º – B) + (180º – C) + 2O = 360°
 B + C = 2O
 180 – A = 2O
1
 O = (180 – A)
2
1
 BOC = 90° – A.
Hence proved.
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PRACTICE EXERCISE
1.
2.
3.
4.
5.
6.
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If an angle is 24° less than its complement. find its measure.
An angle is 40° less than one-third of its supplement. Find the angle and its supplement.
Two supplementary angles are in the ratio 11 : 7. Find them.
Find the angle whose supplement is four times its complement.
Find the measure of an angle if, three times its supplement is 60° more than six times its complement.
In the following figure, it is given that 2a – 5b = 10º find a and b.
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a
A
b
B
7. In the following figures, what value of x will make AOB a straight line?
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8. In thefollowing figure, COA = 90º and AOB is a straight line. Find x and y.
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9. Prove that the bisectors of the angles of a linear pair are at right angles.
10. If the bisectors of two adjacent angles form a right angle then prove that their non-common arms are in
the same straight line.
11. In given figure, AOB is a line. Ray OD is perpendicular to AB . OC is another ray lying between OA and
OD. Prove that DOC =
1
(BOC – AOC)
2
D
B
90°
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A
12. In the following figure, AB and CD intersects at O and BOE = 70°. Find value of a, b and c.
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13. In the given figure 1 : 2 = 5 : 4 and AB || CD. Find all the labelled angles.
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14. Find missing x in the following diagrams. Given AB || CD.
B
A
B
140°
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85°
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E
x°
D
C
(v)
15. In the following figure AB || CD. Find the values of a, b and c.
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16. In the following figure, show that AB || CD.
E
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F
50°
A
30°
20°
C
160°
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D
B
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17. In the following figure, AB || CD and CD || EF. Also EA  AB. If BEF = 65°, find the value of a, b and c.
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x
  x

18. The angles of a triangle are   50  ,   60  and (2x – 15)°, find the angles.
2
 3

19. If a transversal cuts two parallel lines and is perpendicular to one of them, show that it will be perpendicular to the other also.
20. If two parallel lines are intersected by a transversal, show that the bisectors of any corresponding angles
are parallel.
21. If two lines are intersected by a transversal in such a way that the bisectors of a pair of corresponding
angles are parallel, show that the lines are parallel to each other.
—NCERT
22. Prove that the bisectors of a pair of alternate angles of two parallel lines are themselves parallel.
23. Prove that if the arms of an angle are, respectively, parallel to the arms of another angle, then the angles
either have equal measure or they are supplementary.
24. PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the
reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
Prove that AB || CD.
—NCERT
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R
B
Q
D
C
S
25. If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
26. One of the angles of a triangle is 55°. Find the remaining two angles, if their diffrence is 35°.
27. In the given figures find x :
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28. Sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 140° and POT = 105°,
find PRQ.
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29. A square ABCD is surmounted by an equilateral triangle EDC. Find x.
E
x°
D
C
B
A
30. In the following figure, AB || DC. If x 
y
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General Instructions :
y
4
and y  z . Find the values of x, y and z.
2
9
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MM : 30
B
x
z
C
D
A
PRACTICE TEST
Time : 1 hour
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
1. What value of x would make AOB a straight line?
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C
(2x–15)°
x°
(3x + 3)°
A
D
O
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B
MATHEMATICS–IX
2. If two parallel lines are intersected by a transversal, prove that the bisectors of two interior alternate
angles are parallel.
3. Find the value of x°.
4. Find the value of x°, it is given that AB || CD.
B
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5. If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
6. In the given figure, show that AB || CD.
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A
80°
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45°
35°
E
B
C
145°
D
F
7. In the following figure, AB || CD, find a and b.
A
E
B
55°
b
C
a
F
135°
G
D
8. Angles A, B and C of a triangle satisfy B – A = 30° and C – B = 45°. Find all the angles.
9. Prove that the sum of three angles of a triangle is 180°.
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10. In the given figure, the sides AB and AC of ABC are produced to points E and D respectively. If
bisectors BO and CO of CBE and BCD respectively meet at O, then prove that :
BOC = 90° –
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1
BAC
2
A
B
C
E
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D
O
B
ANSWERS OF PRACTICE EXERCISE
1. 33°
2. 15°, 165°
3. 110°, 70°
6. 130°, 50°
7. (i) x = 20° (ii) x = 33° (iii) x = 15°
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5. 20°
8. x = 33°, y = 11°
13. 1 = 3 = 5 = 7 = 80°, 2 = 4 = 6 = 8 = 100°
12. a = 22°, b = 44°, c = 92°
14. (i) x = 115° (ii) x = 40° (iii) x = 55° (iv) x = 80°
15. a = 100°, b = 35°, c = 45°
(v) x = 135°
17. a = 25°, b = 115°, c = 115°
18. 65°, 70°, 45°
26. a = 80°, b = 45°
27. (i) 65° (ii) 95°
29. x = 45°
4. 60°
28. 65°
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30. 24°, 48°, 108°
ANSWERS OF PRACTICE TEST
1. x = 32°
3. x = 110°
7. a = 55°, b = 80°
8. A = 25°, B = 55°, C = 100°
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4. 290°
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