In Fig, PA || BC || DT and AB || DC. Then, the values of a and b are r

1.	In Fig, PA \|\| BC \|\| DT and AB \|\| DC. Then, the values of a and b are respectively(a) 60°, 120° (b) 50°,130° (c) 70°,110° (d) 80°,100°
Answer» (b) 50^o,130^o We know that, ∠PAB = ∠ABC = 50^o … [because interior alternate angles] Given, AB \|\| DC so consider it as parallelogram, In parallelogram adjacent angles of a parallelogram are supplementary. So, ∠ABC + ∠BCD = 180^o 50^o + ∠BCD = 180^o ∠BCD = 180^o – 50^o ∠BCD = 130^o ∠BCD = ∠CDT = 130^o … [because interior alternate angles] Therefore, a = 50^o and b = 130^o

In Fig, PA || BC || DT and AB || DC. Then, the values of a and b are respectively(a) 60°, 120° (b) 50°,130° (c) 70°,110° (d) 80°,100°

Answer»

(b) 50^o,130^o

We know that, ∠PAB = ∠ABC = 50^o … [because interior alternate angles]

Given, AB || DC so consider it as parallelogram,

In parallelogram adjacent angles of a parallelogram are supplementary.

So, ∠ABC + ∠BCD = 180^o

50^o + ∠BCD = 180^o

∠BCD = 180^o – 50^o

∠BCD = 130^o

∠BCD = ∠CDT = 130^o … [because interior alternate angles]

Therefore, a = 50^o and b = 130^o