question_answer

AB || EF, ED || CB and \[\angle \]APE is\[39{}^\circ \], find \[\angle \]CQF.

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 \[=141{}^\circ \]