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八年级数学解答题一般
题目
如图,已知AE⊥BCAE\bot BC,ADAD平分∠BAE\angle BAE,∠ADB=110∘\angle ADB=110^{\circ},∠CAE=20∘\angle CAE=20^{\circ},求∠BAC\angle BAC和∠B\angle B的度数.
知识点：角平分线、平行线的性质章节：未标注

答案与解析

答案

∵AE⊥BC\because AE\bot BC，
∴∠AED=∠AEC=90∘\therefore \angle AED=\angle AEC=90^{\circ}.
∴∠DAE=∠ADB−∠AED=110∘−90∘=20∘\therefore \angle DAE=\angle ADB-\angle AED=110^{\circ}-90^{\circ}=20^{\circ}.
∵AD\because AD平分∠BAE\angle BAE，
∴∠BAE=2∠DAE=40∘\therefore \angle BAE=2\angle DAE=40^{\circ}.
∴∠B=∠AEC−∠BAE=90∘−40∘=50∘\therefore \angle B=\angle AEC-\angle BAE=90^{\circ}-40^{\circ}=50^{\circ}.
∠BAC=∠BAE+∠CAE=40∘+20∘=60∘\angle BAC=\angle BAE+\angle CAE=40^{\circ}+20^{\circ}=60^{\circ}.

解析

∵AE⊥BC\because AE\bot BC，
∴∠AED=∠AEC=90∘\therefore \angle AED=\angle AEC=90^{\circ}.
∴∠DAE=∠ADB−∠AED=110∘−90∘=20∘\therefore \angle DAE=\angle ADB-\angle AED=110^{\circ}-90^{\circ}=20^{\circ}.
∵AD\because AD平分∠BAE\angle BAE，
∴∠BAE=2∠DAE=40∘\therefore \angle BAE=2\angle DAE=40^{\circ}.
∴∠B=∠AEC−∠BAE=90∘−40∘=50∘\therefore \angle B=\angle AEC-\angle BAE=90^{\circ}-40^{\circ}=50^{\circ}.
∠BAC=∠BAE+∠CAE=40∘+20∘=60∘\angle BAC=\angle BAE+\angle CAE=40^{\circ}+20^{\circ}=60^{\circ}.

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