In the figure, the lines GH and IJ are parallel. If ∠1 = 108° and �

1.	In the figure, the lines GH and IJ are parallel. If ∠1 = 108° and ∠2 = 123°, find the value of x, y and z.
Answer» Given GH \|\| IZ ∠1 = 108° ∠2 = 123° ∠1 + ∠KGH = 180 [linear pair] 108° + ∠KGH = 180° 108° + ∠KGH – 108° = 180° – 108° ∠KGH = 72° ∠KGH = x° (corresponding angles if KG is a transversal) ∴ x° = 72° Similarly ∠2 + ∠GHK = 180° (∵ linear pair) 123° + ∠GHK = 180° 123° + ∠GHK – 123° = 180° – 123° ∠GHK = 57° Again ∠GHK = y° (corresponding angles if KH is a transversal) y = 57° x° + y° + z° = 180° (sum of three angles of a triangle is 180°) 72° + 57° + z° = 180° 129° + z° = 180° 129° + z° – 129° = 180° – 129° z = 51° x = 72°, y = 57°, z = 51°

In the figure, the lines GH and IJ are parallel. If ∠1 = 108° and ∠2 = 123°, find the value of x, y and z.

Answer»

Given GH || IZ

∠1 = 108°

∠2 = 123°

∠1 + ∠KGH = 180 [linear pair]

108° + ∠KGH = 180°

108° + ∠KGH – 108° = 180° – 108°

∠KGH = 72°

∠KGH = x° (corresponding angles if KG is a transversal)

∴ x° = 72°

Similarly

∠2 + ∠GHK = 180° (∵ linear pair)

123° + ∠GHK = 180°

123° + ∠GHK – 123° = 180° – 123°

∠GHK = 57°

Again ∠GHK = y° (corresponding angles if KH is a transversal)

y = 57°

x° + y° + z° = 180° (sum of three angles of a triangle is 180°)

72° + 57° + z° = 180°

129° + z° = 180°

129° + z° – 129° = 180° – 129°

z = 51°

x = 72°, y = 57°, z = 51°