Step 1: Draw a line. Mark two points A and B on it so that AB = 6 cm.

Step 2: Using compass with A as center and radius more than half of the length of AB, draw two arcs of same length, one above AB and one below AB.

Step 3: With the same radius and B as center draw two arcs to cut the arcs drawn in step 2. Mark the points of intersection of the arcs as C and D.

Step 4: Join C and D. CD will intersect AB. Mark the point of intersection as O CD is the required perpendicular bisector of AB.
Measure ∠AOC. Measure the length of AO and OB. What do you observe?

We know that \(\angle\)QOR + \(\angle\)ROS + \(\angle\)SOP = 180°
26° + \(\angle\)ROS + 32° = 180°
\(\angle\)ROS + 58° = 180°
Subtracting 58° on both sides
We get, \(\angle\)ROS = 180° – 58° = 122°

(i) Since the angles are linear pair, \(\angle\)ACD + \(\angle\)BCD = 180°
123° + \(\angle\)BCD = 180°
Subtracting 123° on both sides
123° + \(\angle\)BCD – 123° = 180° – 123°
\(\angle\)BCD = 57°
(ii) Since the angles are linear pair, \(\angle\)LNO + \(\angle\)MNO = 180°
46° + \(\angle\)MNO = 180°
Subtracting 46° on both sides
46° + \(\angle\)MNO – 46° = 180° – 46°
\(\angle\)MNO = 134°

\(\angle \)AOC = \(\angle \)AOB + \(\angle \)BOC
= 46° + 51°
= 97°

Yes, In the figure
∠AOB and ∠BOC are adjacent angles.
Also ∠AOB + ∠BOC = 180°
∴ ∠AOB and ∠BOC are supplementary

Let the angles be x and y. Given that one angle exceeds the twice of the other angle by 48^o

y = 2x + 48^o
x + y = 180^o (x and y are interior angles)
x + 2x + 48 = 180^o
3x + 48 = 180^o
3x = 132
x = 44^o
y = 2x + 48^o
= 2 x 44 + 48
= 88 + 48
y = 136

From the figure
60 + 3y - 20 + y + y + 10 = 180.
5y + 50 = 180^o
5y = l30^o
Y = 26^o

Let the angles be 11x and 25x. Since the angles are linear pair.
11x + 25x = 180^o
36x = 180^o
x = 5
The angles are 11x = 11 x 5 = 55^o and
25x = 25 x 5 = 125^o

From the figure
2x + 23 + 3x - 18 = 180^o [linear pair]
5x + 5 = 180^o
5x = 175^o
x = 35^o

The two real - life situations where we use parallel lines are
(i) Railway tracks
(ii) Edges of a door frame

From the figure
∴ x + 3x = 180° (Linear pair of angles)
4x = 180°
x = \(\frac{180°}{4}\)
x  = 45^o
ii) y = 3x (Vertically opposite angles)
y = 3 x 45^o
y = 135^o

Sum of angles at a point = 360°
∴ x + 2x + 3x + 4x + 5x = 360°
15x = 360°
x = \(\frac{360°}{15}\)
x = 24^o
From the given angles the largest angle is 5x
= 5 x 24° = 120°
The largest angle is 120°.

Let the three angles be x, 4x and 7x .
The sum of three angles at a points is supplementary
∴ x + 4x + 7x = 360°
12x = 360°
x^o = \(\frac{{360}^o}{12}\)
x = 30°
If x = 30^o then
4x = 4 x 30° = 120°
7x = 7 x 30° = 210°.
∴ The three angles are 30°, 120° and 210°.

From the figure
∠FEH = ∠FEG + ∠GEH
120° =  34° + ∠GEH
120° - 34° = ∠GEH 
∠GEH = 86°

From the figure the pair ofadjacent angles are
i) \(\angle\)ABG and, \(\angle\)CBG
ii) \(\angle\) DCE and \(\angle\)ECF
iii) \(\angle\)ECE and \(\angle\)ACF
iv) \(\angle\)ACF and \(\angle\)ECD.
v) \(\angle\)BCF and \(\angle\)FCE

Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.

Step 4: Mark the point of intersection as G. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

(i) The angle that corresponds to \(\angle\)1 is \(\angle\)5
(ii) The angle that is alternate interior to \(\angle\)3 is \(\angle\)5
(iii) The angle that is alternate exterior to \(\angle\)8 is \(\angle\)2
(iv) The angle that corresponds to \(\angle\)8 is \(\angle\)4
(v) The angle that is alternate exterior to \(\angle\)7 is \(\angle\)1
(vi) The angle that is alternate interior to \(\angle\)6 is \(\angle\)4