In Fig, line l ∥ m and n is transversal. If ∠1 = 40°, find all th

1.	In Fig, line l ∥ m and n is transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.
Answer» Given that, ∠1 = 40° ∠1 and ∠2 is a linear pair [from the figure] ∠1 + ∠2 = 180° ∠2 = 180° – 40° ∠2 = 140° Again from the figure we can say that ∠2 and ∠6 is a corresponding angle pair So, ∠6 = 140° ∠6 and ∠5 is a linear pair [from the figure] ∠6 + ∠5 = 180° ∠5 = 180° – 140° ∠5 = 40° From the figure we can write as ∠3 and ∠5 are alternative interior angles So, ∠5 = ∠3 = 40° ∠3 and ∠4 is a linear pair ∠3 + ∠4 = 180° ∠4 = 180° – 40° ∠4 = 140° Now, ∠4 and ∠6 are a pair interior angles So, ∠4 = ∠6 = 140° ∠3 and ∠7 are pair of corresponding angles So, ∠3 = ∠7 = 40° Therefore, ∠7 = 40° ∠4 and ∠8 are a pair corresponding angles So, ∠4 = ∠8 = 140° Therefore, ∠8 = 140° Therefore, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40° and ∠8 = 140°

In Fig, line l ∥ m and n is transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.

Answer»

Given that, ∠1 = 40°

∠1 and ∠2 is a linear pair [from the figure]

∠1 + ∠2 = 180°

∠2 = 180° – 40°

∠2 = 140°

Again from the figure we can say that

∠2 and ∠6 is a corresponding angle pair

So, ∠6 = 140°

∠6 and ∠5 is a linear pair [from the figure]

∠6 + ∠5 = 180°

∠5 = 180° – 140°

∠5 = 40°

From the figure we can write as

∠3 and ∠5 are alternative interior angles

So, ∠5 = ∠3 = 40°

∠3 and ∠4 is a linear pair

∠3 + ∠4 = 180°

∠4 = 180° – 40°

∠4 = 140°

Now, ∠4 and ∠6 are a pair interior angles

So, ∠4 = ∠6 = 140°

∠3 and ∠7 are pair of corresponding angles

So, ∠3 = ∠7 = 40°

Therefore, ∠7 = 40°

∠4 and ∠8 are a pair corresponding angles

So, ∠4 = ∠8 = 140°

Therefore, ∠8 = 140°

Therefore, ∠1 = 40°, ∠2 = 140°, ∠3 = 40°, ∠4 = 140°, ∠5 = 40°, ∠6 = 140°, ∠7 = 40° and ∠8 = 140°