In Fig. AB||EF, ED||CB and ∠APE is 39°. Find∠CQF.

1.	In Fig. AB\|\|EF, ED\|\|CB and ∠APE is 39°. Find∠CQF.
Answer» Since ED\|\|BC and AB is a transversal, so ∠QBP = ∠APE [Corresponding angles] or ∠QBP = 39° Now, AB\|\|EF and BC is a transversal. Therefore, ∠FQB = ∠QBP [Alternate interior angles] or ∠FQB = 39° Also, ∠CQF + ∠FQB = 180° [Linear pair] So ∠CQF + 39° = 180° or ∠CQF = 180° – 39° or ∠CQF = 141°

Answer»

Since ED||BC and AB is a transversal,

so ∠QBP = ∠APE [Corresponding angles]

or ∠QBP = 39°

Now, AB||EF and BC is a transversal.

Therefore, ∠FQB = ∠QBP [Alternate interior angles]

or ∠FQB = 39°

Also, ∠CQF + ∠FQB = 180° [Linear pair]

So ∠CQF + 39° = 180°

or ∠CQF = 180° – 39°

or ∠CQF = 141°

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