数学
如图,△ABC的外角∠CBD,∠BCE的平分线相交于点F,若∠A=68°,求∠F的度数
∠F=180°-∠FBC-∠FCB =180°-1/2∠CBD-1/2∠BCE =180°-1/2(180°-∠ABC)-1/2(180°-∠ACB) =1/2(∠ABC+∠ACB) =1/2(180°-∠A) =90°-34° =56°
问:数学△ABC中,∠ABC的角平分线与∠ACB的外角,∠ACD的平分线交于A1 分别计算∠A为70度时,80度时,∠A1的度数?
答:当∠A=70度时,∠A1=35度;当∠A=80度时,∠A1=40度。 因为∠A1=180-(∠A1BC+∠A1CB)且∠A1BC=1/2∠ABC; ...详情>>