In geometry, the types of angles are classified according to the measure. Further we have pairs of angles such as, complementary angles, supplementary angles, adjacent angles and linear pair. Learning these basics thoroughly will help us to learn more concepts on geometry. In the real life examples, we come across adjacent angles frequently like, the angles made by any three fingers, angle made by a door with the wall, the angles made by the vein of a leaf with the midrib etc. In this section let us study about the definition, real life examples and the problems based on the definition of adjacent angles.

Pair of angles which have a common arm and a common vertex, whose interiors do not overlap.

In the above figure, B is the common vertex and BD is the common arm.

The pair of adjacent angles are $\angle ABC$ and $\angle CBD$.

### Adjacent Angles in Real Life

As real life examples we see the adjacent angles in our every life life. The opened door, the angles formed by our fingers, the hands of a clock, the cycle wheels etc are the real life examples of adjacent angles.

The following picture shows the adjacent angles made by an arrow head and the hands of a clock.

Linear Pair: Pair of adjacent angles whose sum is 180o is called a linear pair. In the pair of angles which form a linear pair, there will be a common vertex and a common arm. The non-common arm form a straight angle with the vertex.

In the following figure, the ray PQ lies on the ray RS. Therefore, $\angle RPQ$ and $\angle QPS$ form a linear pair.
(i. e)
$\angle RPQ$ + $\angle QPS$ = 180o

### Solved Examples

Question 1: Find the pair of equal adjacent angles whose sum  is 144o.
Solution:

Let the two equal angles be x.
Therefore, x + x = 144
=>             2x = 144

=>               x = $\frac{144}{2}$

= 72o

Question 2: Two adjacent angles are in the ratio 2 : 7 and the sum of the angles is equal to 54o. Find the angles.
Solution:

Let the angles be 2x and 7x respectively.
Therefore, sum of the angles = 2x + 7x
= 54
=>     9x = 54

=>        x = $\frac{54}{9}$

= 6
The two angles are          2x = 2 ( 6 )
= 12o
and 7x = 7 ( 6 )
= 42o

Question 3: Find the adjacent angles which differ by 30o and the sum of the two angles is equal to 70o.
Solution:

Let the two angles be x and ( x - 30o )
Sum of the angles = x + x - 30
= 70o
=>      2 x - 30 = 70
=>             2 x = 70 + 30
= 100

=>               x = $\frac{100}{2}$

= 50
The other angle will be x - 30 = 50 - 30
= 20o
The two adjacent angles are 20o and 50o.

Question 4: In the following figure, CD $\perp$ AB. Find the measure of x. Hence find the angles, $\angle ACP$ and $\angle CPD$.

Solution:

Since CD is $\perp$ AB, $\angle ACD$ = 90o
As the ray CP divides $\angle ACD$ into pair of adjacent angles,
we have $\angle ACP$ + $\angle PCD$ = 90o
Therefore,                  2x - 20 + 4 x - 10 = 90
=>                           6 x - 30 = 90
=>                                 6 x = 90 + 30 = 120

=>                                    x = $\frac{120}{6}$

= 20
Therefore,                                     $\angle ACP$ = 2 x - 20
= 2 ( 20 ) - 20
= 40 - 20
$\angle ACP$ = 20 o
$\angle PCD$ = 4 x - 10
= 4 ( 20 ) - 10
= 80 - 10
$\angle PCD$   = 70 o

Question 5: In the adjoining figure, AB is a straight line and OC is a ray on AB.
( i ) Find x.
( ii ) Find $\angle AOC$
( iii ) Find $\angle BOC$

Solution:

AOB will be a straight line, if
$\angle AOC$ + $\angle BOC$ = 180 o
Therefore, 3x + 20 + 4x - 36 = 180 o
=>                        7 x - 16 = 180
=>                           7x     = 180 + 16
= 196
=>                             x     = $\frac{196}{7}$
x     = 28 o  ------------------Answer ( i )
$\angle AOC$       = 3 x + 20
= 3 ( 28 ) + 20
= 84 + 20
$\angle AOC$        = 104 o -----------------Answer ( ii )
$\angle BOC$       = 4 x - 36
= 4 ( 28 ) - 36
= 112 - 36
$\angle BOC$      = 76 o ----------------- -Answer ( iii )

Question 6: Pair of adjacent angles on a straight line are in the ratio 3 : 6, find the measure of each angle.
Solution:

Let the angles be 3x and 6x.
As the two angles are on a straight line, the two angles form a linear pair.
Therefore, 3x + 6x = 180
=>                9 x = 180

=>                  x = $\frac{180}{9}$

= 20 o
Hence,          3 x = 3 ( 20 )
= 60 o
and            6 x = 6 ( 20 )
= 120 o
Therefore, the two angles are 60 o and 120 o.

Question 7: Prove that the angle made by the bisectors of pair of angles on a straight line is 90 o.
Solution:

Let the two angles on a straight line be x and y
( i. e ) $\angle AOC$ = x and $\angle BOC$ = y
Therefore, we have x + y = 180 o
Dividing both sides by 2, we get,
$\frac{x}{2}$ + $\frac{y}{2}$ = $\frac{180}{2}$ = 90o  ------------------------- ( 1 )
Since OP is the bisector of  $\angle AOC$ we have $\angle POC$ = $\frac{x}{2}$
Since OQ is the bisector of $\angle COB$, we have $\angle COQ$ = $\frac{y}{2}$
$\angle POC$ + $\angle COQ$ = $\frac{x}{2}$ + $\frac{y}{2}$
= 90 o  [ from ( 1 ) ]
Therefore, the $\angle POQ$ = 90 o
( i. e) the angle between the bisectors is 90 o .

Question 8: The difference between the pair of angles which form linear pair is 90 o. Find the measure of each angle.
Solution:

Let the two angles be x and y which form a linear pair.
Therefore, x + y = 180 o ------------------------ ( 1 )
Since the difference of the two angles = 90 o ,
we have x - y = 90 o --------------------------( 2 )
Adding the equations, ( 1 ) and ( 2 ), we get,
2 x = 270

=>      x = $\frac{270}{2}$

= 135 o
Substituting x = 135 o , in equation ( 1 ), we get,
135 + y = 180
=> y = 180 - 135
=  45 o
Therefore, the pair of angles is 45 0 and 135 o .