Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\) Áp dụng Pytago \(EF=\sqrt{DE^2+DF^2}=25\left(cm\right)\)
Áp dụng HTL:
\(\left\{{}\begin{matrix}DE^2=EH\cdot EF\\DF^2=FH\cdot EF\\DH^2=FH\cdot EH\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}EH=\dfrac{DE^2}{EF}=9\left(cm\right)\\FH=\dfrac{DF^2}{EF}=16\left(cm\right)\\DH=\sqrt{9\cdot16}=12\left(cm\right)\end{matrix}\right.\)
\(b,\sin\widehat{E}=\cos\widehat{F}=\dfrac{DF}{EF}=\dfrac{4}{5}\approx\left\{{}\begin{matrix}\sin53^0\\\cos37^0\end{matrix}\right.\\ \Rightarrow\widehat{E}\approx53^0;\widehat{F}\approx37^0\)
a) \(EF=\sqrt{3^2+4^2}=5\)(cm)
\(DH=\dfrac{DE\cdot DF}{EF}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}=2,4\left(cm\right)\)
b) \(EF=\sqrt{12^2+9^2}=15\left(cm\right)\)
\(DH=\dfrac{DE\cdot DF}{EF}=\dfrac{9\cdot12}{15}=\dfrac{108}{15}=7.2\left(cm\right)\)
c) \(EF=\sqrt{12^2+5^2}=13\left(cm\right)\)
\(DH=\dfrac{DE\cdot DF}{EF}=\dfrac{5\cdot12}{13}=\dfrac{60}{13}\left(cm\right)\)
EH=13-4=9cm
\(DF=\sqrt{4\cdot13}=2\sqrt{13}\left(cm\right)\)
\(DH=\sqrt{4\cdot9}=6\left(cm\right)\)
a) đương nhiên ( áp dụng hệ thức lượng trong tam giác vuông )
b) \(\text{EF}=\sqrt{DE^2+DF^2}=\sqrt{12^2+16^2}=20\) (cm )
ta có DE^2 = EH . EF => EH = DE^2/ EF = 12^2 / 20 = 7.2 ( cm )
DH = DE.DF / EF = 9,6 ( cm )
\(1,\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{\sqrt{5}-2}=\dfrac{2\sqrt{5}-4+2\sqrt{5}+4}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}=4\sqrt{5}\\ 2,\)
a, \(EF=EH+FH=5\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}DE^2=HE\cdot EF=5\\DF^2=HF\cdot EF=20\\DH=FH\cdot EH=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}DE=\sqrt{5}\left(cm\right)\\DF=2\sqrt{5}\left(cm\right)\\DH=2\left(cm\right)\end{matrix}\right.\)
b, \(\sin\widehat{E}=\dfrac{DF}{EF}=\dfrac{2\sqrt{5}}{5};\cos\widehat{E}=\dfrac{DE}{EF}=\dfrac{\sqrt{5}}{5}\)
\(\tan\widehat{E}=\dfrac{DF}{DE}=\dfrac{2\sqrt{5}}{\sqrt{5}}=2;\cot\widehat{E}=\dfrac{1}{\tan\widehat{E}}=\dfrac{1}{2}\)
Xét ΔDEH vuông tại D có đg cao DH
\(FE=HE+HF=1+4=5cm\\ DE^2=EH.FE\\ \Leftrightarrow DE^2=1.5\\ \Leftrightarrow DE=\sqrt{5}cm\\ DF^2=FE^2-DE^2\\ \Leftrightarrow DF^2=5^2-\sqrt{5}^2\\ \Leftrightarrow DF^2=20\\ \Leftrightarrow DF=\sqrt{20}=2\sqrt{5}cm\)
\(EF=EH+FH=1+4=5\left(cm\right)\)
Xét tam giác DEF vuông tại D có đường cao DH ta có:
\(\left\{{}\begin{matrix}DE^2=EH\cdot EF\\DF^2=FH\cdot EF\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}DE=\sqrt{EH\cdot EF}=\sqrt{1\cdot5}=\sqrt{5}\left(cm\right)\\DF=\sqrt{FH\cdot EF}=\sqrt{4\cdot5}=2\sqrt{5}\left(cm\right)\end{matrix}\right.\)
\(a,EF=\sqrt{DE^2+DF^2}=15\left(cm\right)\left(pytago\right)\\ \Rightarrow\sin\widehat{E}=\dfrac{DF}{EF}=\dfrac{9}{15}=\dfrac{3}{5}\\ \cos\widehat{E}=\dfrac{DE}{EF}=\dfrac{12}{15}=\dfrac{4}{5}\\ \tan\widehat{E}=\dfrac{DF}{DE}=\dfrac{9}{12}=\dfrac{3}{4}\\ \cot\widehat{E}=\dfrac{1}{\tan\widehat{E}}=\dfrac{4}{3}\\ b,Áp.dụng.HTL:DH\cdot EF=DE\cdot DF\\ \Rightarrow DH=\dfrac{12\cdot9}{15}=7,2\left(cm\right)\)
