1)Tren cung mot nua mat phang bo chua tia Ox ve tia Oy ; Oz sao cho xOy = 600 ; zOx=1200
a)Chung to Oy la phan giac cua xOz?
b)ve tia Om la tia doi cua tia Ox . Chung to Oz la phan giac cua mOy?
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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
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: \(\widehat{BOy}+\widehat{x'Oy}=\widehat{BOx'}\)
\(\Rightarrow\widehat{x'Oy}=\widehat{BOx'}-\widehat{BOy}=90^o-30^o=60^o\)
\(a)\)
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'
\(\widehat{AOy}+\widehat{x'Oy}=\widehat{AOx'}\)
\(90^o+\widehat{x'Oy}=150^o\)
\(\Rightarrow\widehat{x'Oy}=60^o\)
cho goc aob = 60 độ tren cung mot nua mat phang bo chua tia oa ve tia ox vuong goc voi tia ob . Tren nua mat phang con lai oy vuong goc tia oa
a Chung minh aox = boy
b ve tia ox' la tia doi cua tia ox hãy tinh x'oy
cho goc xoy =50 do. lay diem A tren tia ox . tren cung 1 nua mat phang bo ox chua tia oy, ve tia ot sao cho oy cat nhau tai B va goc oAt =80 do. goi at' la tia phan giac cua goc xAt
a) chung minh At'song song voi Oy
b) tren nua mat phang ko chua diem A bo duong thang oy ve tia Bn song song Ox
ta có\(\widehat{OAt}+\widehat{tAx}=\widehat{OAx}\)
thay\(80^o+\widehat{tAx}=180^o\)
\(\widehat{tAx}=180^o-80^o=100^o\)
vid tia At' là tia phân giác của tAx
\(\Rightarrow\widehat{tAt'}=\widehat{t'Ax}=\frac{\widehat{xAt}}{2}=\frac{100^o}{2}=50^o\)
\(\Rightarrow\widehat{xAt'}=\widehat{xOy}=50^o\)
hai góc \(\widehat{xAt'}\)và\(\widehat{xOy}\)ở vị trí đồng vị bằng nhau
\(\Rightarrow Oy//At'\)
b)
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TREN CUNG MOT NUA MAT PHANG CO BO CHUA TIA Ox VE HAI GOC xOy,yOz KE BU NHAU SAO CHO GOC xOy=80 DO
A] TINH GOC yOz
B] GOI Om , On LA CAC TIA PHAN GIAC CUA GOC xOy VA GOC yOz . CHUNG TO GOC mOm LA GOC VUONG
C] TREN NUA MAT PHANG KO CHUA TIA Oy CO BO CHUA TIA Ox , VE TIA Oc SAO CHO GOC xOt=80 DO . TINH mOt
tren cung mot nua mat phang bo chua tia Ox ve 2 tia Oy va Oz sao cho goc xOy = 80 do va xOz =130 do
tia Oy nam giua 2 tia Ox va Oz ko
tren cung mot nua mat phang co bo chua tia ox ve 2 tia oy oz xoy = 120 xoz =50 tinh yoz goi om la phan giac cua xoz ox' la tia doi cua ox tren nua mp khong chua tia oz có bờ ox ve tia on sao cho xon = 20 hoi oz la tia phan giac cua yon khong vi sao
Cho tia Ox . Ve hai goc xOy va xOz .tren cung mot nua mat phang co bo chua tia Ox sao cho xOy = 50 do , xOz = 65 do
cho hai tia oy,oz cung nam tren mot nua mat phang co bo chua tia ox. biet xoy=30 do,xoz=120 do.tinh so do goc yoz
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