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八年级数学填空题一般
题目
如图,O\odot OABC\triangle ABC的外接圆,BCBC的中垂线与AC^\widehat {AC}相交于DD点.若B=74\angle B=74^{\circ},C=46\angle C=46^{\circ},则AD^\widehat {AD}的度数为______.
知识点:线段垂直平分线的性质、全等三角形的性质、全等三角形的判定、勾股定理、直角三角形的性质、圆周角定理I、切线的判定、相似三角形的性质I、相似三角形的判定I、相似三角形的判定与性质章节:未标注

答案与解析

答案

B=74\because \angle B=74^{\circ}C=46\angle C=46^{\circ}
A=60\therefore \angle A=60^{\circ}
BC^=120°\therefore \widehat {BC}=120°AB^=92°\widehat {AB}=92°
BDC^=240°\therefore \widehat {BDC}=240°
BC\because BC的中垂线与AC^\widehat {AC}相交于DD点,
BD^=12BDC^=120°\therefore \widehat {BD}=\frac{1}{2}\widehat {BDC}=120°
AD^=BD^AB^=120°92°=28°\therefore \widehat {AD}=\widehat {BD}-\widehat {AB}=120°-92°=28°
故答案为:2828^{\circ}.

解析

B=74\because \angle B=74^{\circ}C=46\angle C=46^{\circ}
A=60\therefore \angle A=60^{\circ}
BC^=120°\therefore \widehat {BC}=120°AB^=92°\widehat {AB}=92°
BDC^=240°\therefore \widehat {BDC}=240°
BC\because BC的中垂线与AC^\widehat {AC}相交于DD点,
BD^=12BDC^=120°\therefore \widehat {BD}=\frac{1}{2}\widehat {BDC}=120°
AD^=BD^AB^=120°92°=28°\therefore \widehat {AD}=\widehat {BD}-\widehat {AB}=120°-92°=28°
故答案为:2828^{\circ}.

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