In each of the following figures AB || CD. Find the unknown angles, giving reasons.

(i)

(ii)

(i)

Given:

AB || CD

Since, AB is a straight line:

x° + 100° = 180° $\quad$[Linear pair]

⇒ x° = 180° - 100°

⇒ x° = 80°

y° = 120° $\quad$[Vertically opposite angles]

AB || CD and GH is a transversal:

t° = x° $\quad$[Alternate angles]

∴ t° = 80°

Let the angle opposite to z° be p°.

In quadrilateral ABCD, sum of interior angles is 360°.

∴ 120° + 100° + 80° + p° = 360°

⇒ 300° + p° = 360°

⇒ p° = 360° - 300°

⇒ p° = 60°

z° = p° $\quad$[Alternate angles]

∴ z° = 60°

x° = 80°, y° = 120°, z° = 60° and t° = 80°

(ii)

Given:

AB || CD

t° = 120° $\quad$[Vertically opposite angles]

AB || CD and EF is a transversal:

y° = t° $\quad$[Alternate angles]

∴ y° = 120°

x° = 50° $\quad$[Exterior alternate angles]

Since, AB is a straight line:

x° + z° = 180° $\quad$[Linear pair]

⇒ 50° + z° = 180°

⇒ z° = 180° - 50°

⇒ z° = 130°

x° = 50°, y° = 120°, z° = 130° and t° = 120°