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,\(\angle EFC=80^0\)

Since ED II BC and AB is a transversal, so \(\angle QBP=\angle APE\)       [corresponding angles]
\(\angle QBP=39^0\)
Now, AB II EF and BC is a transversal.
Therefore, \(\angle FQB=\angle QBP\) [alternate interior angles]
\(\angle FQB=39^0\)
Also,  \(\angle CQF+\angle FQB=180^0\) [linear pair]
so \(\angle CQF=39^0=180^0\)
or \(\angle CQF=180^0-39^0\)
or \(\angle CQF=141^0\)

Since points A, O and B are collinear (Given), therefore AB is a straight line.
(a) As O is a point on the line AB, therefore
\(\angle AOD+\angle DOC+\angle BOC=180^0\)
\(\angle AOD+\angle BOC+90^0=180^0\)
\(\angle AOD+\angle BOC=90^0\)
\(\angle AOD\ \)and \(\angle BOC\)are complementary angles.
(b) Also, \(\angle AOC\) and \( \angle BOC\)are supplementary as \(\angle AOC+ \angle BOC=180^0\)

Since PQ II RS and QR is transversal.So
x = 39° [alternate interior angles]
Again QR II ST and RS is a transversal. 
Therefore, y = x [alternate interior angles]
or y = 39°

In the given figure (v), \(\angle \) 1 and \(\angle \)2 are adjacent angles because they have a common vertex and a common arm but no common interior point.

In the given figure \(\angle \) 1 and \(\angle \)2 are adjacent angles because they have a common vertex and a common arm but no common interior point.

The supplement angle of 55° is 125°

Let the supplement angle of 100° be x°,
We know that, the sum of two supplementary angles is 180°.
\(\therefore\) x° + 100°= 180° \(\Rightarrow\)  x° = 180° -100° = 80°
Hence, the supplement angle of 100° is 80°.

In this pair, measures of the given angles are 45° and 45°.
\(\therefore\) Sum of the given angles = 45° + 45° = 90°,
which is less than 180°.
So, this pair of angles is not supplementary.

In this pair, sum of two angles =35° + 55° = 90°
which is equal to 90°.
So, this pair of angles is complementary.

In this pair, sum of two angles = 75° + 25° = 100° 
which is greater than 90°.
So, this pair of angles is not complementary.

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°.

Let the complement angle of 45° be x°.
We know that, the sum of two complementary angles is 90°.
\(\therefore\) x°+ 45° = 90°
\(\Rightarrow\) x° = 90° - 45° = 45°
[transposing 45° to RHS]
Hence, the complement angle of 45° is 45°.

No, two obtuse angles can never be complement to each other because their sum cannot be 90°, as an obtuse angle has its measure greater than 90°.
e.g. 105° and 135° are two obtuse angles.
Their sum =105° + 135° = 240° \(\neq \)  90°
Therefore, two obtuse angles cannot be complement to each other.