耘绮
幼苗
共回答了18个问题采纳率:83.3% 举报
【方法一】
做AD⊥BC于D:
AD=ABsin30°=15*1/2=15/2
BD=ABcos30°=15*根号3/2=15根号3 /2
CD = 根号(AC^2-AD)^2 = 根号{10^2-(15/2)^2} = 5根号7 /2
当D点在BC上时:
BC = BD+CD = 15根号3 /2 + 5根号7 /2 = (15根号3+5根号7) /2
当D点在BC延长线上时:
BC = BD-CD = 15根号3 /2 - 5根号7 /2 = (15根号3-5根号7) /2
【方法二】
根据余弦定理:
AC^2=AB^2+BC^2-2*AB*BC*cosB
10^2 = 15^2 + BC^2 -2*15*BC*cos30°
BC^2 - 15根号3 BC + 125 = 0
(BC-15根号3/2)^2 = (15根号3/2)^2-125 = 175/4
BC-15根号3/2 = ±5根号7/2
BC = (15根号3+5根号7) /2 ,或BC = (15根号3-5根号7) /2
1年前
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