6.3 Transversals and Parallel Lines

Assumption: Standard NCERT Fig 6.29 configuration. AB || CD || EF. x and y are co-interior angles between AB and CD (Wait, usually x and y are consecutive?). Let's define clearly: x is angle on AB, y on CD, z on EF. x and y are on same side? No, standard problem is x and y are co-interior. y and z are alternate? Let's solve generic logic.

Setup: AB || CD || EF. Transversal cuts. $x$ and $y$ are on the same side between AB and CD ($x+y=180$). $y$ and $z$ are on opposite sides? No, usually angles are defined such that we relate them.

Correct Logic for standard problem: AB || EF. $x$ and $z$ are Alternate Interior Angles. So $x = z$.

Given $y : z = 3 : 7$. Let $y = 3k, z = 7k$.

Between CD and EF: $y$ and $z$ are co-interior sum 180? Or Alternate? If they are on the same side, $y+z=180$. If alternate, $y=z$.

Let's assume $y$ and $z$ are co-interior (angles on same side of transversal). Then $3k + 7k = 180 \implies 10k = 180 \implies k=18$. $z = 7(18) = 126^\circ$. Since $x=z$ (Alternate interior from AB to EF), $x=126^\circ$.

Alternative interpretation (NCERT Ex 6.2 Q1): $x+y=180$ (co-interior). But relation is between y and z? Usually $x=z$ is the key.