Sum of two angles = 90° + 80° = 170°,
which is less than 180°.
So, this pair is not a linear pair.

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.

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, sum of two angles = 48° + 52° = 100°
which is greater than 90°.
So, this pair of angles is not complementary.

The complement angle of 41° is 49°

∵ \(\angle \) 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

Since, QP||RS, PR is a transversal line.
\(\therefore \quad \angle QPR=\angle PRS=65°\) [alternate angles]
\(\angle a=65°\) \(\left[ \angle \quad PRS=a \right] \)
\(\angle PQS=\angle b=70°\) [corresponding angles]
Hence, \(\angle \)a and \(\angle \)b are 65° and 70°, respectively.

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°​​​​​​​

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 complement be 90° - x°,
Since, the angle is equal to its complement.
x°=90° - x°
On transposing  from RHS to LHS, we get
x° + x° = 90° \(\Rightarrow\) 2x° = 90°
On dividing both sides by 2, we get
\(\frac { 2x }{ 2 } =\frac { 90° }{ 2 } \Rightarrow \quad x=45°\)
Hence, the required angle is 45°.

Given, AB II ML, \(\angle \)LOB =142°
\(\angle \)A = 32°
\(\angle \)MOA and \(\angle \)A are alternate angles.
So, \(\angle \)MOA = 32°
Also, (\(\angle \)MOA + x) and \(\angle \)142° form a linear pair. So,
\(\angle \)MOA + x + 142° =180°
32° + x + 142° = 180°
x =180° -142° - 32°
x = 180° -174°
x = 6°
Anil's answer in the test was not correct. The value of x was 6°.
Hence, Anil does not understand the concept of alternate angles and linear pair. So, he was still confused about the lines and angles chapter.

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°