Ppt

1. MODULE :

2. Lines and Angles
B
A
C
A + B + C = 180
o

3. PAIRS OF ANGLES
 VERTICALLY OPPOSITE ANGLES
A
B
C
D
AOC and BOD,
O
form pairs of vertically opposite angles.
AOD and COB
Note : Vertically Opposite angles are EQUAL
AOC = BOD, AOD = COB

4. MODULE :

5. D
B
C
A
 They have common vertex.
 They have common ray.
 Non-common arms should lie on
either side of common arm.
Two angles are said to be adjacent
angles if
ADB, BDC
D D
Have common vertex ‘D’.
Have common arm ray BD.
ray AD and ray DC are non
common arm lying on either
side of common arm BD

6.  They have common vertex.
 They have common ray.
 Non-common arms should lie on
either side of common arm.
Two angles are said to be adjacent
angles if
Consider PQR and SQR
Q Q
QR is the common ray.
Non-common arms QS and QP lie on the
same side of common arm QR
R
S
Q
P
Have common vertex Q.
PQR and SQR are not adjacent angles.

7. MODULE :

8. PAIRS OF ANGLES
 LINEAR PAIR ANGLES
Two adjacent angles are said to form a linear pair angle,
 If their non-common rays
A
C
O
B
A
AOB and BOC are the
angles in a linear pair.
form a pair of opposite rays.

9. PAIRS OF ANGLES
 LINEAR PAIR ANGLES
The sum of the measure of a linear pair angle is 180°
C
O
B
A
AOB + BOC = 180°

10. MODULE :

11. If a ray stands on a line, then the sum of the two adjacent
angles so formed is 1800 and vice–versa.
1. If, OC stands on line AB
then,
2. If
mAOC + mCOB = 1800
then,
AB is a line and OC stands on the line.
i.e. A , O , B are collinear
AOC COB
+ = 1800
AOC COB
+ = 1800
O B
A
C

12. A B
C D
AB  CD
P Q
PQ  CD
If
AB  CD
and
PQ  CD
then
AB  PQ
Lines which are parallel to a given line are parallel
to each other.

13. MODULE :

14. In figure, lines AB and CD intersect at O. If
A B
O
C
D
E
AOC+BOE =70ºand  BOD= 40º,find BOE
and reflex  COE
Sol: AOC
But,
AOC
 = 400
BOE
 = 700 [ From (1) and (2) ]
= 300
COE
 = 1800
+ BOE = 700
BOE – AOC
= 700
[Vertically opposite angles]
….(1)
[ BOD = 400 ]
∵
….(3)
– 400
BOE

BOE

+ BOD

+
COE
 = 1800
300
+ 400
+
COE
 = 1100
COE
 = 1800 400
–
O 40º
AOC = BOD
Reflex of COE = 360
?
40º 30º
….(2)
– COE
300
–
Ex . : 6.1 - 1

15. In figure, lines AB and CD intersect at O. If
A B
O
C
D
E
AOC+BOE =70ºand  BOD= 40º,find BOE
COE
 = 1100
Reflex COE
 = 250º
40º
40º
30º
Reflex COE
 = 360º – 110–
and reflex  COE
Sol:
Ex . : 6.1 - 1