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如图,CE⊥AF,垂足为E,CE与BF交于点D,∠F=50°,∠C=30°,求∠EDF和∠DBA的度数。
参考答案:40°,70°解:\(\because CE \bot AF\),\(\therefore \angle FED = 90^\circ \),\(\because \angle F = 50^\circ \),\(\therefore \angle EDF = 90^\circ - \angle F = 90^\circ - 50^\circ = 40^\circ \),\(\therefore \angle CDB = \angle EDF = 40^\circ \),\(\because \angle C = 30^\circ \),\(\therefore \angle DBA = \angle C + \angle CDB = 30^\circ + 40^\circ = 70^\circ \),即\(\angle EDF = 40^\circ \),\(\angle DBA = 70^\circ \)。