In the given figure, l ∥ m and t is a transversal. If ∠1 and ∠2

1.	In the given figure, l ∥ m and t is a transversal. If ∠1 and ∠2 are in the ratio 5: 7, find the measure of each of the angles ∠1, ∠2, ∠3 and ∠8.
Answer» Two parallel lines l and m be cut by a transversal t, forming angles. It is given that ∠1: ∠2 = 5: 7 Let the measures of angel be 5x and 7x, Then, 5x + 7x = 180° 12x = 180° x = 180/12 x = 15° ∴ ∠1 = 5x = 5 × 15 = 75° ∠2 = 7x = 7 × 15 = 105° We know that, ∠2 + ∠3 = 180° … [∵ linear pair] 105°+ ∠3 = 180° = ∠3 = 180° – 105° = ∠3 = 75° ∠3 + ∠6 = 180° … [∵ the sum of the consecutive interior angle is 180°] 75° + ∠6 = 180° ∠6 = 180 – 75 ∠6 = 105° Now ∠6 = ∠8 = 105° … [∵ vertically opposite angles are equal] ∴ ∠1= 75°, ∠2 = 105°, ∠3 = 75° and ∠8 = 105°.

In the given figure, l ∥ m and t is a transversal. If ∠1 and ∠2 are in the ratio 5: 7, find the measure of each of the angles ∠1, ∠2, ∠3 and ∠8.

Answer»

Two parallel lines l and m be cut by a transversal t, forming angles.

It is given that ∠1: ∠2 = 5: 7

Let the measures of angel be 5x and 7x,

Then,

5x + 7x = 180°

12x = 180°

x = 180/12

x = 15°

∴ ∠1 = 5x = 5 × 15 = 75°

∠2 = 7x = 7 × 15 = 105°

We know that,

∠2 + ∠3 = 180° … [∵ linear pair]

105°+ ∠3 = 180°

= ∠3 = 180° – 105°

= ∠3 = 75°

∠3 + ∠6 = 180° … [∵ the sum of the consecutive interior angle is 180°]

75° + ∠6 = 180°

∠6 = 180 – 75

∠6 = 105°

Now ∠6 = ∠8 = 105° … [∵ vertically opposite angles are equal]

∴ ∠1= 75°, ∠2 = 105°, ∠3 = 75° and ∠8 = 105°.