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![](https://rs.olm.vn/images/avt/0.png?1311)
góc F=90-30=60 độ
Xét ΔDEF vuông tại D có sin E=DF/EF
=>DF/20=1/2
=>DF=10cm
=>DE=10*căn 3(cm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,EF=\sqrt{DE^2+DF^2}=15\left(cm\right)\left(pytago\right)\\ \Rightarrow\sin\widehat{E}=\dfrac{DF}{EF}=\dfrac{9}{15}=\dfrac{3}{5}\\ \cos\widehat{E}=\dfrac{DE}{EF}=\dfrac{12}{15}=\dfrac{4}{5}\\ \tan\widehat{E}=\dfrac{DF}{DE}=\dfrac{9}{12}=\dfrac{3}{4}\\ \cot\widehat{E}=\dfrac{1}{\tan\widehat{E}}=\dfrac{4}{3}\\ b,Áp.dụng.HTL:DH\cdot EF=DE\cdot DF\\ \Rightarrow DH=\dfrac{12\cdot9}{15}=7,2\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{DF}{EF}=\dfrac{4}{5}\)
\(\Leftrightarrow DF=\dfrac{4}{5}EF\)
\(\Leftrightarrow DF=24\left(cm\right)\)
\(\Leftrightarrow FE=30\left(cm\right)\)
\(\Leftrightarrow DI=14.4\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn tự vẽ hình giúp mik nha
vẽ đường cao EH (H\(\in\)DF)
ta có: \(\widehat{F}\)=180\(^o\)-\(\widehat{E}\)-\(\widehat{F}\)=180-70-60=50
EH=EF.sinF=30.sin50=22,98
sinD=\(\dfrac{EH}{ED}\)\(\Rightarrow\)ED=\(\dfrac{EH}{sinD}\)=\(\dfrac{22,98}{sin60}\)=26,54
DH=\(\sqrt{DE^2-EH^2}\)(pytago)=\(\sqrt{26,54^2-22,98^2}\)=13,28
HF=\(\sqrt{EF^2-EH^2}\)(pytago)=\(\sqrt{30^2-22,98^2}\)=19,29
mà:DF=DH+HF=13,28+19,29=32,57
chu vi \(_{\Delta DEF}\)=DE+EF+DF=26,54+30+32,57=89,11
\(S_{\Delta DEF}\)=\(\dfrac{EH.DF}{2}\)=\(\dfrac{22,98.32,57}{2}\)=374,2293