In Fig, AB and CD are parallel lines intersected by a transversal by a

1.	In Fig, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.
Answer» Given that, ∠LMD = 35° From the figure we can write ∠LMD and ∠LMC is a linear pair ∠LMD + ∠LMC = 180°[sum of angles in linear pair = 180°] On rearranging, we get = ∠LMC = 180° – 35° = 145° So, ∠LMC = ∠PLA = 145° And, ∠LMC = ∠MLB = 145° ∠MLB and ∠ALM is a linear pair ∠MLB + ∠ALM = 180° [sum of angles in linear pair = 180°] = ∠ALM = 180° – 145° = ∠ALM = 35° Therefore, ∠ALM = 35°, ∠PLA = 145°.

In Fig, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.

Answer»

Given that, ∠LMD = 35°

From the figure we can write

∠LMD and ∠LMC is a linear pair

∠LMD + ∠LMC = 180°[sum of angles in linear pair = 180°]

On rearranging, we get

= ∠LMC = 180° – 35°

= 145°

So, ∠LMC = ∠PLA = 145°

And, ∠LMC = ∠MLB = 145°

∠MLB and ∠ALM is a linear pair

∠MLB + ∠ALM = 180° [sum of angles in linear pair = 180°]

= ∠ALM = 180° – 145°

= ∠ALM = 35°

Therefore, ∠ALM = 35°, ∠PLA = 145°.