From the figure,
∠YCX = 48°
∠YCX = LACB (Vertically opposite angles)
ஃLACB = 48°
The exterior angle LCBD = Sum of two interior opposite angles.
= ∠BAC + ∠ACB
= 57° + 48° = 105°
∠CBD = ∠CBE + ∠EBD
105° = 65° + ∠EBD
∠EBD = 105°-65° = 40° 
The exterior angle ∠BDZ = The two interior opposite angles.
∠BED + ∠EBD
y = 97° + 40° = 137°
y = 137°

48^o + x + <FGD = 180^o
<FGD = 180^o - 48^o - x
<FGD = 132^o - x
By Exterior angle property
<FGD = x + 68^o
From (1) and (2)
132^o - x = x + 68
132^o - 68^o = x + x
64^o = 2x
x = 32^o
<FGD = x + 68^o
= 32^o + 68^o
= 100^o

By Exterior angle property
x + 28^o = 62^o
x = 62^o - 28^o = 34^o
Now in ABC
x + y + 28^o = 180^o
34^o + y + 28^o = 180^o
62^o + y = 180^o
y = 180^o - 62^o = 118^o
x = 34^o, y = 118^o

BC = BD
BE = AB
<CBA = <DBE
By SAS criterion \(\triangle C D B \cong \angle E B D\)
The two sides AC is parallel to DE.

\(\triangle M N O \cong \triangle D E F\)
Corresponding angles
<M = <D = 60^o
<N = <E = 45^o
<M + <N + <O = 180^o
60^o + 45^o + <O= 180^o
105^o + <O = 180^o
<O = 180^o - 105^o = 75^o

(i)


Steps:
1. Draw a line segmentAB = 7 cm.
2. At A, draw a ray AX making an angle of 120⁰ with AB.
3. With A as centre, draw an arc of radius 6.5 em to cut the ray AX at C.
4. Join BC.   
(ii)

Steps:
1. Draw a line segment \(\overline { BC } =8cm\)
2. At e draw a ray ex making an angle of radius.
3. With e as centre draw an arc of radius 6 cm.
4. Join AB.
(iii)

Steps:
1. Draw a line segment AB = 5 cm.
2. At B, draw a ray BX making an angle of 110° with AB.
3. With B as centre, draw an arc of radius 5 em to cut the ray BX at C.
4. Join AC.

congruent triangles by AAA

Criterion: ASA

ㄥB and ㄥE are Not corresponding angles

Corresponding sides
\(\overline{P Q} \text { matches } \overline{M N}\)
\(\overline{Q R} \text { matches } \overline{L M}\)
\(\overline{P R} \text { matches } \overline{L N}\)
Congruent angles
<P = <N
<R = <L
<Q = <M

(i) Corresponding Congruent Sides
\(\overline{A B}=\overline{D E} \)
\(\overline{A C}=\overline{D F} \)
\(\overline{B C}=\overline{E F} \)
(ii) Corresponding Congruent Angles
\(\angle A=\angle D \)
\(\angle B=\angle E \)
\(\angle C=\angle F \)

Exterior angle = Sum of two interior opposite angles
6x - 5 = 100 - x + 2x
6x - 5 = 100 + x = 6x - x = 100 + 5
5x = 105
\(x=\frac{105}{5}=21^{\circ}\)

In  ΔABC 
Let \(\angle\)A = x
\(\angle\)B = 3 times \(\angle\)A
\(\angle\)B = 3 x
\(\angle\)C = 2 times \(\angle\)A
= 2x
\(\angle A+\angle B+\angle C={ 180 }^{ o }\) 
x + 2x + 3x = 180^o 
6x = 180^o
\(x=\frac{180^{\circ}}{6}=30^{\circ}\)
\(\angle A={ 30 }^{ o }\) 
\(\angle B=3\angle A=3\times { 30 }^{ o }={ 90 }^{ o }\) 
\(\angle C=2\angle A=2\times { 30 }^{ o }={ 60 }^{ o }\) 
The three angles are 30°,60°,90°

Let the three angles are 3x, 5x and 4x
= 3x, 5x, 4x = 180°
12x = 180°
\(x=\frac{180^{\circ}}{12}=15^{\circ}\)
3x = 3 x 15°= 45°
5x = 5 x 15°= 75°
4x = 4x 15°= 60°
The three angles are 45°,75°, 60°

\(\angle A+\angle N+\angle G+\angle L+\angle E+\angle S\)
\(=\underbrace{\angle A+\angle E+\angle G}_{\text {Sum of three angles of } \Delta}+\underbrace{\angle S+\angle N+\angle L}_{\text {Sttm of three angles of } \Delta}\)
= 180^o + 180^o = 360^o