Let \(\angle EFC=x\)
Then \(\angle AFD=x\)
It is given that CD intersects line AB at F.
Therefore,\(\angle CFB=\angle AFD\) (vertically opposite angles)
So,x=50⁰
But \(\angle EFA=\angle AFD\) which gives \(\angle EFA=50^0\)
Now, \(\angle CFB+\angle EFA+\angle EFC=180^0\) [as AB is a straight line]
\(50^0+50^0+\angle EFC=180^0\)
\(\angle EFC=180^0-100^0\)
Thus,\(\angle EFC=80^0\)

Let x be the complement of an angle, then the angle will be \(\frac { 2 }{ 3 } \)x.
We know that, sum of complementary angle is 90°
So, \(x+\frac { 2 }{ 3 } x=90°\quad \Rightarrow \frac { 3x+2x }{ 3 } =90°\)
\(\Rightarrow \quad 5\times =270°\quad \Rightarrow x=\frac { 270° }{ 5 } =54°\Rightarrow x=54°\)
\(\therefore \quad \frac { 2 }{ 3 } \times { 54 }^{ ° }=2\times 18°=36°\)
Hence, complement angle of 54° is 36°

Since, n is a transversal to l and m.
\(\therefore\) \(\angle \)1 = 75°
[ vertically opposite angles]
Also, \(\angle \)1 = 75° = 75° + 75° = 150°
\(\therefore\) 150° \(\neq \) 180°
So, l and m are not parallel.

\(\angle \)3 and \(\angle \)4 are alternate interior angles

Equal supplementary angles are \(\angle \)BOE and \(\angle \)EOD

We know that, the sum of two complementary angles is 90°. So, if an angle is greater than 45°, then its complementary angle would be less than 45°.

\(\angle \)BOD and \(\angle \)BOC are not adjacent angles because their other arms OD and OC are not on the opposite of the common arm OB, as shown in adjoining figure.

In this pair, measures of the given angles are 50° and 130°.
\(\therefore\) Sum of the given angles = 50° + 130° = 180°,
So, this pair of angles is supplementary.

In this pair, sum of two angles = 70° + 20° = 90° So, this pair of angles is complementary.

Let the complement angle of 65° be x°.
We know that, the sum of two complementary angles is 90°.
\(\therefore\) x° + 65° = 90° \(\Rightarrow\) x°= 90° - 65° = 25°
[transposing 65° to RHS]
Hence, the complement angle of 65° is 25°.

x + 2x = 180°     (Co-interior angle)
3x = 180°
x = 60°

∵ \(\angle \) BAD = \(\angle \)BAE + \(\angle \)EAD
= 40° + 30° = 70°
and \(\angle \)CDA = 7·0°
∴\(\angle \)BAD = \(\angle \)CDA
But they form a pair of alternate angles.
=> AB || CD ...(1)
Also, \(\angle \)BAE + \(\angle \)AEF = 40° + 140° = 180°
But they form a pair of interior opposite angles.
=>AB || EF
From (1) and (2), we get
AB || CD || EF
=> CD || EF

a + 60° = 180°
\(\Rightarrow \) a = 180° - 60° = 120°
a = d = 120° [alternate exterior angles]
b + d = 180° [linear pair]
\(\Rightarrow \) b + 120° = 180°
\(\Rightarrow \) b = 180° - 120° = 60°
\(\Rightarrow \) c = b = 60°. [vertically opp. angles.]

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, lines I and m are parallel to each other, where line t is intersecting both these lines at the point 0 and M, respectively.
So, \(\angle \)OMA and \(\angle \)BME are vertically opposite angles.
So, \(\angle \)OMA = 42°
Also, \(\angle \)OMA and \(\angle \)x are interior angles.
So, \(\angle \)x + \(\angle \)OMA =180° \(\Rightarrow\) \(\angle \)x = 180° - 42° = 138°

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]
\(\Rightarrow 75^0=30^0+\angle b\)
\(\Rightarrow 75^0-30^0=\angle b\)
\(\Rightarrow \angle b=45^0\)

Given,\(\angle 1=(2a+b)^0\) and  \(\angle 6=(3a-b)^0\)
so, \(\angle 1=\angle 7\) [alternate exterior angles]
So, we have \(\angle 7=(2a+b)^0\)
\(\therefore \angle 6+\angle 7=180^0\)              [linear pair]
\(\Rightarrow\) (3a-b)⁰ + (2a + b)⁰ = 180°
\(\Rightarrow\) 3a- b + 2a + b = 180°
\(\Rightarrow\) 5a = 180°
on dividing both sides by 5, we get
\(\Rightarrow a=\frac{180^0}{5}=36^0\)
\(\because \angle 1+\angle 2=180^0\)                    [linear pair]
\(\Rightarrow 2a+b+\angle 2=180^0\)
\(\Rightarrow 2\times36^0+b+\angle 2=180^0\)
\( \Rightarrow b+\angle 2=180^0-72^0\)
\(\Rightarrow b+\angle 2=180^0\)
\(\therefore \angle 2=(108-b)\)

75, 75, 60

Given, \(\angle \)PSR = 115°; \(\angle \)RQD = 70° and \(\angle \)CPF = 65°
\(\angle \) PQR + \(\angle \)RQD = 180°   [linear pair]
\(\Rightarrow\) \(\angle \)PQR = 180° - 70°  \(\Rightarrow\)  \(\angle \)PQR = 110°
\(\angle \)CPF = \(\angle \)SPQ = 65 ​​​​​°
[vertically opposite angles]
If EF || GH
Then, according to the definition of cointerior angles, the sum of \(\angle \)SPQ + \(\angle \)RQP should be 180°.
\(\therefore\) 65° + 110°​​​​​​​ = 175°​​​​​​​ \(\neq \) 180 ​​​​​°​​​​​​​
​​​​​​​So, EF is not parallel to GH.

Given, p || q and I is a transversal.
We know that, the sum of pair of interior angles on the same sides of the transversal is supplementary.
\(\therefore\) \(\angle \)a + 60° = 180° \(\Rightarrow\) \(\angle \)a = \(\angle \)180° - 60° = 120°
and I || m and q is a transversal.
\(\therefore\) \(\angle \)a = \(\angle \)1          [pair of corresponding angles]

\(\angle \)1 = 120°
and \(\angle \)d= \(\angle \)1= 120°    [vertically opposite angles]
Now, \(\angle \)1 + \(\angle \)c= 180°
[\(\because\) sum of the angles on the same side of a transversal is 180°]
\(\Rightarrow\) 120° + \(\angle \)c = 180° \(\Rightarrow\) \(\angle \)c = 180° -120° = 60°
\(\Rightarrow\) \(\angle \)b = \(\angle \)c = 60° [vertically opposite angles]
Hence, \(\angle \)a = 120°, \(\angle \)b = 60°, \(\angle \)c = 60° and \(\angle \)d=120°

\(\angle 1=\angle 2;\angle 2+\angle 2=100^\circ\)
\(\therefore \angle 1=50^\circ\)
\(\therefore \angle 4\) = 180° - 50° = 130°.

\(\angle 2=\angle 6\), corresponding angles.

\(\angle x-\angle y\) = (180° - 62°) - 62°
                 = 118° - 62° = 56°.

Definition of a linear pair of angles.

180° - 90° = 90°.

90° - 45° = 45°.