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 \(\angle AOC+ \angle BOC=180^0\)

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

The complement angle of 41° is 49°

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.]

In the given figure \(\angle AOB\) and \(\angle COB\) are the angles of linear pair.
So, \(\angle AOB+\angle BOC=180^0\)
(3x + 10°) + (2x - 30°) = 180°
\(\Rightarrow\) 3x + 10° + 2x - 30° = 180°
\(\Rightarrow\) 5x - 20° = 180°
\(\Rightarrow\) 5x = 180° + 20°
\(\Rightarrow\) 5x = 200°
\(\Rightarrow x=\frac{200^0}{5}\)
Thus, x=40⁰
Now, \(\angle AOB=3x+10^0\)
= 3 (40°) + 10°
= 120° + 10°
= 130°.​​​​​​​

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

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°