Question

[Hoàng Trọng Chính( ɻɛɑm ʙáo cáo )]

Cho AOB=60⁰. Tren cung mot nua mat phang bo OB chua tia OA, ve tia Ox vuong goc voi tia OB. Tren nua mat phang kia, ve tia Oy vuong goc voi tia OA

a) Chung minh: AOx = BOy

b) Ve Ox^, la tia doi cua tia Ox. Hay tinh x^,Oy

Answer 1

[.]

a) Trên cùng 1 nửa mặt phẳng bờ OB chứa tia OA, vẽ tia Ox và OA sao cho \(\widehat{BOx}=90^o;\widehat{AOB}=60^o\)

\(\Rightarrow\widehat{AOB}>\widehat{BOx}\left(90^o>60^o\right)\)

\(\Rightarrow\) Tia OA nằm giữa 2 tia OB, Ox

Ta có: \(\widehat{AOB}+\widehat{AOx}=\widehat{BOx}\)

\(\Rightarrow\widehat{AOx}=\widehat{BOx}-\widehat{AOB}=90^o-60^o=30^o\)

Vì tia OB nằm giữa 2 tia OA, Oy nên: \(\widehat{AOy}=\widehat{AOB}+\widehat{BOy}\)

\(\Rightarrow\widehat{BOy}=\widehat{AOy}-\widehat{AOB}=90^o-60^o=30^o\)

\(\Rightarrow\widehat{AOx}=\widehat{BOy}\)   (đpcm)

b) Vì tia Oy nằm giữa 2 tia OB, Ox' nên ta có: \(\widehat{BOy}+\widehat{x'Oy}=\widehat{BOx'}\)

\(\Rightarrow\widehat{x'Oy}=\widehat{BOx'}-\widehat{BOy}=90^o-30^o=60^o\)

Answer 2

[Ngô Chi Lan]

\(a)\)

Vì \(\widehat{AOB}< \widehat{BOx}\left(60^o< 90^o\right)\)nên tia OA nằm giữa hai tia Ox và OB

\(\widehat{AOB}+\widehat{AOx}=\widehat{BOx}\)

\(60^o+\widehat{AOx}=90^o\)

\(\Rightarrow\widehat{AOx}=30^o\)

Vì \(\widehat{AOB}< \widehat{AOy}\left(60^o< 90^o\right)\)nên tia OB nằm giữa hai tia Oy và OA

\(\widehat{AOB}+\widehat{BOy}=\widehat{AOy}\)

\(60^o+\widehat{BOy}=90^o\)

\(\Rightarrow\widehat{BOy}=30^o\)

\(\Rightarrow\widehat{AOx}=\widehat{BOy}=30^o\)

\(b)\)

Vì \(\widehat{AOx}< \widehat{xOx'}\left(30^o< 180^o\right)\)nên tia OA nằm giữa hai tia Ox và Ox'

\(\widehat{AOx}+\widehat{AOx'}=\widehat{xOx'}\)

\(30^o+\widehat{AOx'}=180^o\)

\(\Rightarrow\widehat{AOx'}=150^o\)

Vì \(\widehat{AOy}< \widehat{AOx'}\left(90^o< 150^o\right)\)nên tia Oy nằm giữa hai tia OA và Ox'

\(\widehat{AOy}+\widehat{x'Oy}=\widehat{AOx'}\)

\(90^o+\widehat{x'Oy}=150^o\)

\(\Rightarrow\widehat{x'Oy}=60^o\)