Answer:
Since ED || BC and AB is a transversal, so So \[\angle QBP\text{ }=\text{ }\angle APE\] [Corresponding angles] or \[\angle QBP\text{ }=\text{ }39{}^\circ \] Now, AB | | EF and BC is a transversal. Therefore, ZFQB = ZQBP [Alternate interior angles] or \[\angle FQB\text{ }=\text{ }39{}^\circ \] Also, \[\angle \]CQF +\[\angle \]FQB \[=180{}^\circ \] [linear pair] so \[\angle \]CQF + 39° \[=180{}^\circ \] or \[\angle \]CQF \[=180{}^\circ -39{}^\circ \] or \[\angle \]CQF \[=141{}^\circ \]
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