Hình dưới cho biết $d // d' // d''$ và hai góc $60^{o}, 110^{o}$. Tính các góc $\widehat{E_{1}}, \widehat{G_{2}}, \widehat{G_{3}}, \widehat{D_{4}}, \widehat{A_{5}}, \widehat{B_{6}}$.
A5C6Bdd'd"G231ED60*110*4
$d' //d''\Rightarrow \widehat{E_1}=\widehat{C}=60 (slt);\widehat{G_2}=\widehat{BDd'}=110(đv)$
$\widehat{G_2}+\widehat{G_3}=180$ (kề bù ) $\Rightarrow \widehat{G_3}=180-\widehat{G_2}=180-110=70$
$d'//d''\Rightarrow \widehat{D_4}=\widehat{G_2}=110(slt)$
$d//d''\Rightarrow \widehat{E_1} = \widehat{A_5}=60(đv); \widehat{B_6}=\widehat{G_3}=70(đv)$

CÓ +)$d'//d''\Rightarrow \widehat{E_{1}}=\widehat{C}=60^o$(2 góc so le trong)

+)$d'//d''\Rightarrow \widehat{G_{2}}=\widehat{D}=110^o$(2 góc đồng vị)

+)$\widehat{G_{3}}+\widehat{G_{2}}=180^o$(2 góc kề nhau)
$\Leftrightarrow \widehat{G_{3}}=180^o-\widehat{G_{2}}=180^o-110^o=70^o$

+)$\widehat{D_{4}}=110^o$(2 góc đối đỉnh)

+)$\widehat{ACD}=\widehat{C}=60^o$(2 góc đối đỉnh)
$d//d'\Rightarrow \widehat{A_{5}}=\widehat{ACD}=60^o$(2 góc so le trong)

+)$d//d''\Rightarrow \widehat{B_{6}}=\widehat{G_{3}}=70^o$
nhầm đến v là cùng –  Kaito kid 25-08-16 08:46 PM
nhầm :3 . –  Lionel Messi 25-08-16 08:41 PM
cảm ơn ô –  Lionel Messi 25-08-16 08:41 PM
tán gái nhiều quá ak mà 180-110=50 =='' –  Kaito kid 25-08-16 08:40 PM

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