Given, l || m || n and p || q
\(\therefore \quad \angle 6a=120°\) [corresponding angles]
\(\Rightarrow \quad \angle a=\frac { 120° }{ 6 } =20°\)
Similarly, \(\angle\)4c=120°    [corresponding angles]
\(\Rightarrow \quad \angle c=\frac { 120° }{ 4 } =30°\quad and\quad \angle 3b=\angle 4c\)
[ corresponding angles]
\(\Rightarrow \quad \angle b=\frac { 4\times 30° }{ 3 } \Rightarrow b=40°\)

Given, \(\angle \)2 = 58° \(\Rightarrow\) \(\angle \)1 = \(\angle \)2    [alternate angles]
Also, \(\angle \)1 + \(\angle \)3 =180°    [linear pair]
\(\therefore\) 58°+ \(\angle \)3=180° \(\Rightarrow\) \(\angle \)3=180°-58°
\(\Rightarrow\) \(\angle \)3 =122°
So, \(\angle \)1 = 58° and \(\angle \)3 = 122°

Since, PQ||RT and RQ is a transversal line.
\(\angle \)RPQ and \(\angle \)a are corresponding angles.
\(\therefore \quad \angle RPQ=\angle a\Rightarrow \angle a=45°\)
Also, \(\angle \)b and \(\angle \) RQP are alternate angles.
\(\therefore\)   \(\angle \)b= 55°
Hence, \(\angle \)a + \(\angle \) b = 45° + 55°​​​​​​​ =100°

Since, \(\angle \)b and \(\angle \)132° are pair of interior angles on the same side of transversal line.
\(\therefore\) \(\angle \)b + 132° =180° [\(\because\) cointerior angles]
\(\Rightarrow\) \(\angle \)b=180°-132°\(\Rightarrow\)  \(\angle \)b=48°
Also, \(\angle \)a + \(\angle \)b and \(\angle \)65° are pair of interior angles
on the same side of transversal line.
\(\therefore\) \(\angle \)a + \(\angle \)b + 65° =180° [by linear pair]
\(\Rightarrow\) \(\angle \)a + 48° + 65° =180°
\(\Rightarrow\) \(\angle \)a=180°-113°  \(\Rightarrow\)\(\angle \)a=67°
Hence, \(\angle \)a=67° and \(\angle \)b=48°

Since, angles \(\angle\)3x and \(\angle\)2x form a linear pair.
So, 3x° +2x°  = 180°  \(\Rightarrow\) 5x° = 180°
\(\Rightarrow \quad { x }^{ ° }=\frac { 180° }{ 5 } =36°\)
So, 3x = 3 \(\times\)36° = 108°
and 2x = 3 X 36° = 72°

Since, lines l₁ and l₂ are parallel to each other.
where l₃ is transversal
\(\therefore\)   \(\angle \) P and \(\angle \)130° form a linear pair angles.
So,  \(\angle \)P+130°=180°
\(\Rightarrow\) \(\angle \)P = 180° - 130°
\(\Rightarrow\) \(\angle \)P = 50°
Also, \(\angle \)P and \(\angle \)M, alternate interior angles.
\(\angle \)M=\(\angle \)P=50°

Since, lines l₂ and l₃ are parallel to each other,
where l₁ is transversal line.
\(\therefore\) \(\angle \)60° and m are alternate interior angles.
So, m = 60°
Also, angles m and n form a linear pair angles.
So, m + n = 180 ° \(\Rightarrow\) 60 ° + n = 180 °
\(\Rightarrow\) n = 180°-60°​​​​​​​ \(\Rightarrow\) n = 120°​​​​​​​

Since, angles 46° and x are vertically opposite angles.
So, x = 46°
\(\because\) Angles x and y form a linear pair angles.
\(\therefore\) x+y=180° and x = 46°
So, 46°+y=180° \(\Rightarrow\) y = 180°​​​​​​​-46°​​​​​​​
\(\Rightarrow\) y = 134°​​​​​​​
Now, angles y and z are vertically opposite angles.
​​​​​​​So, y = z = 134°​​​​​​​

Since, angles 40° and y° are linear pair angles.
\(\therefore\) 40° +y° = 180°
\(\Rightarrow\) y = 180° -40°  \(\Rightarrow\) y = 140°
Again, angles 140° and x are linear pair angles.
\(\therefore\) 140° +x = 180°
\(\Rightarrow\) x = 180° -140°  \(\Rightarrow\) x = 40° ​​​​​​

Given, p II q
\(\angle \)e + 125° = 180°  [by linear pair]
\(\Rightarrow\) \(\angle \)e = 180° -125° \(\Rightarrow\)\(\angle \)e = 55°
\(\therefore\) \(\angle \)f = \(\angle \)e = 55°      [vertically opposite angles]
Since, p II q and t is a transversal.
\(\therefore\) \(\angle \)a = \(\angle \)f = 55° [alternate interior angles]
\(\angle \)d = 125° [corresponding angles]
\(\angle \)c = \(\angle \)a = 55° [vertically opposite angles]
and \(\angle \)b = \(\angle \)d = 125° [vertically opposite angles]
Hence, \(\angle \)a = 55°, \(\angle \)b = 125°, \(\angle \)c = 55°, \(\angle \)d = 125°, \(\angle \)e = 55° and \(\angle \)f = 55°.

Let ABC be an equilateral triangle.
Since, all the angles of an equilateral triangle are equal.
\(\therefore\) \(\angle \)A = \(\angle \)B = \(\angle \)C = x° [say]
We know that, sum of all the angles of a triangle is 180°.
\(\therefore\) \(\angle \)A + \(\angle \)B + \(\angle \)C = 180°
\(\Rightarrow \) x° +x° +x° = 180°
\(\Rightarrow \) 3x° ​​​​​​=180°
\(\Rightarrow \quad x°=\frac { 180° }{ 3 } =60°\)
Hence, \(\angle \)A = \(\angle \)B =\(\angle \)C= 60°​​​​​​​
​​​​​​​

Let the angle be x°,
Therefore, its supplement be 180°- x°,
Since, the angle is equal to its supplement.
\(\therefore\) x°=180°-x°
On transposing x° from RHS to LHS, we get
x°+ x°= 180° \(\Rightarrow\) 2x° = 180°
On dividing both sides by 2, we get
\(\therefore \quad { x }^{ ° }=\frac { 180° }{ 2 } =90°\)
Hence, the required angle is 90°.

Given, \(\angle \)1= 30°
Since, \(\angle \)1 and  \(\angle \)2 form a linear pair.
\(\therefore\) \(\angle \)1+ \(\angle \)2 = 180° \(\Rightarrow\) 30° + \(\angle \)2 = 180°
[transposing 30° to RHS]
\(\Rightarrow\)  \(\angle \)2 = 180° - 30°\(\Rightarrow\) \(\angle \) 2 = 150°
Now, \(\angle \)3 = \(\angle \)1= 30° [vertically opposite angles]
Hence, \(\angle \)2 = 150° and \(\angle \)3 = 30°

Let the smaller angle be x°,
Therefore, its supplement larger angle = (x + 44)°
Since, the sum of two supplement angles is 180°.
\(\therefore \) x° + (x + 44)° = 180°
\(\Rightarrow \) x° + x° + 44° = 180° \(\Rightarrow \) 2x° + 44° = 180°
On transposing 44° from LHS to RHS, we get
2x° = 180° - 44° \(\Rightarrow \) 2x° = 136°
On dividing both sides by 2 ,we get \(\frac { 2x° }{ 2 } =\frac { 136° }{ 2 } \) = 68°
\(\Rightarrow \) x° = 68°
Hence, the smaller angle is 68° and its supplement larger angle =68° + 44° = 112°

Let one angle be x°
Therefore, other complement angle be 90° - x°.
[\(\because\) sum of two complementary angles is 90°]
Given, difference of two complementary angles = 12°
\(\therefore\) x°-(90°-x°)=12°
\(\Rightarrow \) x° - 90° + x°= 12°
\(\Rightarrow \)2x°= 12° + 90° \(\Rightarrow \) 2x° = 102°
[on transposing (- 90°) from LHS to RHS]
On dividing both sides by 2, we get
\(\frac { { 2x }^{ ° } }{ 2 } =\frac { 102 }{ 2 } \Rightarrow { x }^{ ° }=51°\)
\(\therefore\) One angle = 51°
and its complement angle = 90°-51°​​​​​​​ = 39°​​​​​​​
Hence, the required complementary angles are 51°​​​​​​​ and 39°​​​​​​​​​​​​​​

Given, PQ II RS II UT
(a) Given, c = 57° and \(a=\frac{c}{3}\)
\(\because PQ||UT\)
\(\therefore \angle UTP=\angle QPT\) [alternate interior angles]
\(\Rightarrow \angle c=\angle a+\angle b[\because QPT=a+b]\)
\(\Rightarrow 57^0=\frac{57^0}{3}+\angle b\)
\(\Rightarrow 57^0-19^0=\angle b\)
\(\Rightarrow \angle b=38^0\)
\(\therefore \angle b+\angle d=180^0\)
\(\Rightarrow \angle d=180^0-38^0=142^0\)
(b) Given, c = 75° and \(a=\frac{2}{3}c\)
\(\Rightarrow c=75^0\) and \(a=\frac{2}{5}\times75^0=30^0\)
\(\therefore \angle c=\angle a+\angle b\) [alternate interior angles]
\(\Rightarrow 75^0=30^0+\angle b\)
\(\Rightarrow 75^0-30^0=\angle b\)
\(\Rightarrow \angle b=45^0\)

Angle between band c = 30°
[vertically opposite angles]

Since, \(\angle\)AOB = 90° [right angle at O]
So, \(\angle\)AOB = \(\angle\)COB + \(\angle\)COA
\(\because\) \(\angle\)COB = 49° \(\Rightarrow\) \(\angle\)COA + 49° = 90° [given]
\(\Rightarrow\) \(\angle\)COA = 90° - 49° \(\Rightarrow\) \(\angle\)COA = 41°
Since, DC is a straight line, where \(\angle\)COA and \(\angle\)AOD form a linear pair .
So, \(\angle\)AOD + \(\angle\)COA =180°
\(\because\) \(\angle\)COA =41°
\(\therefore\)  \(\angle\)AOD =180° - 41° \(\Rightarrow\)  \(\angle\)AOD =139°