6.2 PAIRS OF ANGLES:
If the sum of two angles is
900 — Complementary angles.
1800 —– Supplementary angles.
Two angles are said to be adjacent angles if
( i ) They have the same vertex 
( ii ) They have a common arm
( iii ) The non common arms are on either side of the arm.
Here ∠ AOB and ∠ BOC are adjacent angles.
Two angles will form a linear pair if their sum is 180
∠ POQ + ∠ POR = 180°
Here ∠POQ and ∠POR are the two linear angles.
When two lines intersect each other the vertically
opposite angles are formed.
Vertically opposite angles are equal to each other.
( i ) ∠ ADC and ∠ BOD
( ii ) ∠ AOD and ∠ BOC
are vertically opposite angles and they are equal
Exercise 6.1
1 ) In Fig 6.13, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°.
Find ∠ BOE and reflex ∠ COE
Given : ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°
To find : ( i ) reflex ∠ COE ( ii ) ∠ BOE
Proof: (i)Since AB is a straight line
∠ AOC + ∠ COE + ∠ BOE = 180°
∠ 70o + ∠ COE = 180°
∴ ∠ COE = 180° − 70° ⇒ ∠ COE = 110°
∴ Reflex ∠ COE = 360° – 110° = 250° ⇒ Reflex ∠ COE = 250°
( ii ) ∠ AOC = ∠ BOD = 40° ( vertically opposite angles )
∴ ∠ AOC = 40°
But ∠ AOC + ∠ BOE = 70° ( given )
400 + ∠ BOE = 70° ⇒ ∠ BOE = 70° − 40° = 30°
∴ ∠ BOE = 30°
2 ) In Fig. 6.14, lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3. Find c
Given : ∠ POY = 90°
a : b = 2 : 3
To find : the value of c
Proof: Let the multiplicative ratio be x
∴ a = 2 x and b = 3 x
a + b = 90° ( given that ∠ POY = 90° )
2 x + 3 x = 90° ⇒ 5 x = 90° ⇒ x = 180°
∴ a = 2 x = 2 x 18 = 360 ⇒ a = 360°
b = 3 x = 3 x 18 = 54° ⇒ b = 54°
Now ( b + c ) form a linear pair
∴ b + c = 180° ⇒ 54° + c = 180°
c = 180° – 54° = 126° ⇒ c = 126°
3 ) In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.
Given: ∠ PQR = ∠ PRQ
To prove : ∠ PQS = ∠ PRT
Proof: Since ST is a straight line
∠ PQS + ∠ PQR = 180° ……. (i)
and ∠ PRT + ∠ PRQ = 180° ………….(ii)
From ( i ) and ( ii ) we get
∠ PQS + ∠ PQR = ∠ PRT + ∠ PRQ
( But ∠ PQR = ∠ PRQ is given )
∴ ∠ PQS = ∠ PRT.
4 ) In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Given: x + y = w + z
To prove: AOB is a line.
Proof: Sum of all the angles around a point is 360°
x, y, w, z are angles around the point O
∴ ∠ x + ∠ y + ∠ z + ∠ w = 360° – – – – – – – ( 1 )
and ( x + y ) = ( z + w ) ( given )
Put this value in ( 1 ) we get
( x + y ) + ( x + y ) = 360° ⇒ 2 x + 2 y = 360°
2 ( x + y ) = 360° ⇒ x + y = 180°
∴ AOB is a straight line.
5 ) In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ.
OS is another ray lying between rays OP and OR. Prove that
∠ ROS = 1/2 ( ∠ QOS − ∠ POS )
Given : POQ is a line (∠ ROQ = ∠ ROP = 90° )
To prove : ∠ ROS = 1/2 ( ∠ QOS – ∠ POS
Proof : Since POQ is a straight line
∠ POS + ∠ ROS + ∠ ROQ = 180° ⇒ ∠ POS + ∠ ROS + 90° = 180°
∴ ∠ POS + ∠ ROS = 90° – – – – – – – – ( 1 )
We also have
∠ ROS + ∠ ROQ = ∠ QOS ⇒ ∠ ROS + 90° = ∠ QOS
∠ QOS − ∠ ROS = 90° – – – – – – – – – – – ( 2 )
From ( 1 ) and ( 2 ) we get
∠ POS + ∠ ROS = ∠ QOS − ∠ ROS
∠ ROS + ∠ ROS = ∠ QOS − ∠ POS
2 ∠ ROS = ∠ QOS − ∠ POS
∠ ROS = 1/2 ( ∠ QOS − ∠ POS )
6 ) 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
Given : ∠XYZ = 64° and ray YQ bisects ∠ ZYP
To prove: ∠ XYQ = ?
Proof : If ray YQ bisects ∠ ZYP we get
∠ ZYQ = ∠ QYP = x
( i ) We know that XYP is a straight line
∠ XYZ + ∠ ZYQ + ∠ QYP = 180°
64° + x + x = 180° ⇒ 64° + 2 x = 180°
2 x = 116° ⇒ x = 58°
∴ ∠ ZYQ = ∠ QYP = 58°
Now ∠ XYZ + ∠ ZYQ = ∠ XYQ
64° + 58° = ∠ XYQ
∴ ∠ XYQ = 122°
( ii ) Reflex ∠ QYP = 360° − ∠ QYP
= 360° − 58°
= 302°
∴ Reflex ∠ QYP = 302°