AB=AC,AB⊥AC

来源:学生作业帮助网 编辑:作业帮 时间:2024/12/01 05:15:56
AB=AC,AB⊥AC
xRn@*R%*,{e*hncXv%*A"6@-!~Jq/p=N"e>sÈF׽0nmcɤmm'(H#{=Rlshu[Yt;h&鼳ehevUZQdmBw`@@9FS2l6^`Z8v)s'Lj^G9XIv*/‹S ;uTl0hf ;z|4٘&~3<-n?bR8"sus]ѩ{49Mys6xQl8a;sl8VPg;9< pp`r S8o5f@p{lcтؚB}jѮ5n3'wÍK%xڭ^SV%ƀx"m (@1

AB=AC,AB⊥AC
AB=AC,AB⊥AC

AB=AC,AB⊥AC
AC=AB,外加垂直条件得出∠ABC=∠ACB=45°,∠BAC=90°
梯形上下底平行,∠ABC+BAD=180°计算得∠BAD=135°
设AB=AC=1,根据勾股定理得BC=BD=√2
根据正弦定理
AB/sin(∠ADB)=BD/sin(∠BAD)
1/sin(∠ADB)=√2/sin(135°)=√2/(√2/2)=2
得出sin(∠ADB)=0.5查表可得∠ADB=30°梯形上下底平行∠ADB=∠DBC=30°
∠DBC=30°,BC=BD,∠DCB=∠CDB=75°