1.
Ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
Mà \(\widehat{B}=\widehat{C}\)
\(\Rightarrow\widehat{A}+\widehat{C}+\widehat{C}=180^0\)
\(\widehat{A}=180^0-2.65^0\)
\(\widehat{A}=50^0\)
2.
Áp dụng định lý pitago, ta có:
\(DF^2=DE^2+EF^2\)
\(\Rightarrow EF=\sqrt{DF^2-DE^2}=\sqrt{17^2-8^2}=\sqrt{225}=15cm\)
Ta có:
\(DF>EF>DE\)
\(\Rightarrow\widehat{E}>\widehat{D}>\widehat{F}\)
Đúng 3
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1.
Ta có:
ˆA+ˆB+ˆC=1800A^+B^+C^=1800
Mà ˆB=ˆCB^=C^
⇒ˆA+ˆC+ˆC=1800⇒A^+C^+C^=1800
ˆA=1800−2.650A^=1800−2.650
ˆA=500A^=500
2.
Áp dụng định lý pitago, ta có:
DF2=DE2+EF2DF2=DE2+EF2
⇒EF=√DF2−DE2=√172−82=√225=15cm⇒EF=DF2−DE2=172−82=225=15cm
Ta có:
DF>EF>DEDF>EF>DE
⇒ˆE>ˆD>ˆF
Đúng 3
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