From this figure
<1 = 180^o - 65^o
x + 39^o = 115^o
x = 115^o - 39^o
x = 76^o
<2 = 65^o
2x - 3y = 65^o
2 x 76 - 3y = 65^o
152 - 3y = 65^o
3y = 152^o - 65^o
3y = 87^o
y = 29^o

From the figure
<1 = 30^o and <2 = 80^o [Alternate interior  angles]

(i) 3a = 126°
a = \(\frac{126°}{3} \) = 42°
a = 42°
(ii) 4a + 13^o + 135° = 180°
4a + 148° = 180°
4a = 180° - 148°
4a = 32°
a = \(\frac{32°}{4}\)
a = 8°
(iii)  8a + 29° = 45°
8a = 45° - 29^o
8a = 16°
a = 2°
(iv) 6a = 90°
a = \(\frac{32°}{4}\) 
a = -15°

From all the figures, all the pair of angles are alternate interior (or) exterior angle.
i) y = 28^o
ii) y = 58^o
iii) y = 123^o
iv) y = 108^o

(a) 60° c
Construction:
Step 1 : Drawn the given angle ∠ABC with the measure 60° using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given ∠ABC
Now ∠ABG = ∠CBG = 30°

(b) 100°
Construction:
Step 1: Drawn the given angle ∠ABC with the measure 100° using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as centre drawn an arc in the interior of ∠ABC and another arc of the same measure with centre at F to cut the previous arc.
Step 4: Marked the point of intersection at G. Drawn a ray BX through G. BG is the required bisector of angle ∠ABC
∠ABG = ∠GBC = 50°

(c) 90°
Construction :
Step 1: Drawn the given angle ∠ABC with the measure 90° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC
Step 3: With the same radius and E as center drawn 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 interaction as G. Drawn a ray BX through G BG is the required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 45°

(d) 48°
Construction:
Step 1: Drawn the given angle ∠ABC with the measure 48° using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as E on BA and F on BC.
Step 3: With the same radius and E as center drawn an arc in the interior of ∠ABC and another arc of the same measure with center at F to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
Now ∠ABC = ∠GBC = 24°

(e) 110°
Construction:
Step 1: Drawn the given angle ∠ABC with the measure 110° using protractor
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked points of intersection as E on BA and F BC.
Step 3: With the same radius and E as center, drawn 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. Drawn a ray BX through G. BG is the required bisector of the given angle ∠ABC
∠ABG = ∠GBC = 55°.