Question

  Giải hộ

Answer 1

a) Ta có: \(a//Gx\)\(\Rightarrow\widehat{aEG}=\widehat{G_1}\) (hai góc so le trong)

\(\Rightarrow\widehat{G_1}=42^o\) (vì \(\widehat{aEG}=42^o\))

b) Có: \(\widehat{GFy}+\widehat{GFb}=180^o\) (hai góc kề bù)

\(\Rightarrow138^o+\widehat{GFb}=180^o\) (vì \(\widehat{GFy}=138^o\))

\(\Rightarrow\widehat{GFb}=180^o-138^o=42^o\)

c) Ta thấy: \(Gx//b\)\(\Leftrightarrow\widehat{G_2}=\widehat{GFb}=42^o\) (hai góc so le trong)

Lại có: \(\widehat{EGF}=\widehat{G_1}+\widehat{G_2}\) (hai góc kề nhau)

\(\Rightarrow\widehat{EGF}=42^o+42^o=84^o\left(cmt\right)\)

#Urushi

Answer 2

a) Do Gx // a

⇒ ∠G₁ = ∠GEa = 42⁰ (so le trong)

b) Ta có:

∠GFb + ∠GFy = 180⁰ (kề bù)

⇒ ∠GFb = 180⁰ - ∠GFy

= 180⁰ - 138⁰

= 42⁰

c) Do Gx // b

⇒ ∠G₂ = ∠GFb = 42⁰ (so le trong)

⇒ ∠EGF = ∠G₁ + ∠G₂

= 42⁰ + 42⁰

= 84⁰

Answer 3

a) \(\widehat{G_1}\) = 42^o

b) \(\widehat{GFb}\) = 42^o

c) \(\widehat{EGF}\) = 84^o