Page 5 - Ans Sub

Construction: PQ is produced to X which intersect AB at Y.
PQ || RS ⇒ PX || RS

∵ AB || CD and PX is transversal
∠CQP + ∠AYX = 180° [Co-exterior angles]
⇒ 60° + ∠AYX = 180°
⇒∠AYX = 180° – 60°
⇒∠AYX = 120°
∵ PX || RS and AB is transversal
∠YRS = ∠AYX [Corresponding angles]
⇒∠YRS = 120° …(i)
and AB || CD and RQ is transversal
∠YRQ = ∠RQD [Alternate interior angles]
⇒∠YRQ = 25° …(ii)
Now, ∠SRQ = ∠SRY + ∠YRQ
⇒∠SRQ = 120° + 25° [From (i) and (ii)]
⇒∠SRQ = 145°

7. In the figure, AB || CD and BC || ED. Find the value of ‘x’ .

Solution:
From the given figure, we know that
AB and CD are parallel line and BC is a transversal
We know that ∠BCD and ∠ABC are alternate angles
So,
0
∠BCD + ∠ABC = x
We also know that BC || ED and CD is a transversal.
From the figure, we know that ∠BCD and ∠EDC form a linear pair of angles
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