135°

281°

In this figure, they have common vertex O, common arm OA and non-common arms OB and OC lying on the either side of OA.
\(\therefore\) \(\angle \)AOB (i.e. \(\angle \)2) and \(\angle \)AOC (i.e. \(\angle \)1) are adjacent angles and they have common vertex O, common arm OD and non-common arms DC and OB.
So, \(\angle \)COD (i.e. \(\angle \)4) and \(\angle \)DOB (i.e. \(\angle \)3) are adjacent angles.

Similarly \(\angle \)3, \(\angle \)2 and \(\angle \)4, \(\angle \)1 are adjacent angles.

Given, a II d, b II e, c II f, a, b, c, d, e and f are line segments.
Give them points A, B, C, 0, E, F, G, H, I, J and K.

\(\because\) \(\angle\)AKE =120°
\(\therefore\)\(\angle\)JKL = \(\angle\)AKE = 120°
[vertically opposite angles)
Now, a ll d
\(\therefore\) \(\angle\)JKL + \(\angle\)KCH=180°          [cointerior angles]
\(\Rightarrow\) 120° + \(\angle\)z = 180°
\(\Rightarrow\) \(\angle\)z = 180° - 120° = 60°
Also, \(\angle\)FGH + \(\angle\)KCH = 180°   [cointerior angles]
\(\Rightarrow\) LFGH = 180° - 60° = 120°
\(\angle\)FGH= \(\angle\)BGD      [vertically opposite angles]
\(\Rightarrow\) \(\angle\)BGD =120° \(\Rightarrow\) \(\angle\)y=120°
and \(\angle\)x + \(\angle\)FGH = 180°  [cointerior angles]
\(\therefore\) \(\angle\)x = 180°-120° = 60°
Hence, \(\angle\)x = 60°, \(\angle\)y = 120° and \(\angle\)z = 60°.

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.

\(\angle \)CPF + \(\angle \)CPS = 180°  [linear pair]
\(\Rightarrow\) 65°​​​​​​​ + \(\angle \)CPS = 180°​​​​​​​
\(\Rightarrow\)  \(\angle \)CPS = 180° - 65° = 115°
If AB || CD, then according to the definition of alternate angles,
\(\angle \)RSP = \(\angle \)CPS
It is satisfying the condition.
​​​​​​​So, AB is parallel to CD.

Since, lines EF and GH are parallel to each other.
Where, \(\angle\)EAB = 70° and \(\angle\)ACH =120°
\(\angle\)EAB and \(\angle\)CBA are alternate angles.
So, \(\angle\)CBA = 70°
\(\because\) (\(\angle\)EAB + \(\angle\)BAC) and \(\angle\)ACH are alternate angles.
so, \(\angle\)EAB + \(\angle\)BAC = 120°
\(\Rightarrow\) 70° + \(\angle\)BAC =120°
 \(\angle\)BAC =120° \(\Rightarrow\)-70° = 50°
\(\because\)\(\angle\)CAE = \(\angle\)EAB + \(\angle\)BAC
\(\therefore\) \(\angle\)CAE = 70° + 50° = 120°
Also, \(\angle\)FAC + \(\angle\)CAE =180°
\(\Rightarrow\)\(\angle\)FAC = 180° -120°
\(\therefore\) \(\angle\)FAC = 60°

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

Since, AF II ED and EF is a transversal line.
So, \(\angle \)AFE = \(\angle \)FED = 40° [alternate interior angles]
Also, AB II CD
70° + (\(\angle \)AFE + \(\angle \)EFD) form a linear pair angles

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°