Question

如图,在Rt△ABC中,∠BAC=90°,∠C=40°,AD是BC边上的高,BE平分∠ABC,AD,AE相交于

如图,在Rt△ABC中,∠BAC=90°,∠C=40°,AD是BC边上的高,BE平分∠ABC,AD,AE相交于点P，求∠DAC和∠AEB的度数

Answer 1

∠DAC=90-40=50 ∠AEB=90-50/2=75

Answer 2

因为 ∠C=40°
∠CBA=50;
BE平分∠ABC
∠EBC=25
∠AEB=90-∠EBC=65
又因为 AD垂直BC
∠DAC=90-∠C=90-40=50

Answer 3

http://video.baidu.com/v?word=%CE%DE%B7%A8%CC%D3%C0%EB%B5%C4%B1%B3%C5%D1&ct=301989888&rn=20&pn=0&db=0&s=0&fbl=800

Answer 4

∠DAC=∠ADC－∠C=90º－40º=50º
∠AEB=∠BAC－½∠ABC=90º－50º÷2=65º

Answer 5

∠DAC=180-∠C-∠ADC=50
∠AEC=90-1/2×50=65