Giúp mình với

xoy và yoz phụ nhau = 180*

ta có xoy + yoz = 180*

=> xoy + 35* = 180*

=> xoy = 180* - 35 *

=> xoy = 145 *

Vì góc xOy và yOz là 2 góc phụ nhau 

=> góc xOy + góc yOz = 90⁰

Thay số: xOy + 35⁰ = 90⁰

=> góc xOy = 90⁰ - 35⁰ = 55⁰

(Vì máy mình không thể viết được kí hiệu góc nên khi làm bài bạn lưu ý nhé!)

Đã là bài hình thì bn vẽ hình ra là sẽ làm đc ngay bn mến nhé

Minh da ve hinh nhung khong giai duoc phan b nen ban giai ho minh nhe

a)Do 2 góc xoy và yoz kề bù =>Hai tia Ox và Oz là hai tia đối nhau

và  xoy + yoz =180°,mà xoy=70°

=> 70° +yoz=180°=>yoz=110°

Vậy yoz =110°

b)Do tia Ot là tia phân giác của góc xoy=> Tia Ot nằm giữa hai tia Ox và Oy (1)

và yot=1/2xoy=1/2.70=35°

Do tia Om là tia phân giác của yoz=>Tia om nằm giữa hai tia Oy và Oz         (2)

và yom=1/2yoz=1/2.110=55°

Từ (1) và (2)=> Tia Oy nằm giữa hai tia Ot và Om

=>toy+yom=tom=>35°+55°=mot

=>mot=90°

Vậy mot=90°

 °

thank you bạn nào trả lời trước cho .... với câu trả lời hợp lí !

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2 goc bu nhau chu

ok mình sai bạn giải chi mình đi

a) ta có 2 góc xOy và x'Oy là 2 góc kề bù nên :

góc yOx + góc yOx' = 180 độ 

=> 45 độ + góc yOx' = 180 độ

                 góc yOx' = 180 độ - 45 độ

                 góc yOx' = 135 độ

b)mik ko bít cách tính chỉ bít kết quả thui . kết quả : tOx' = 45 độ

c) tia Ot' ko thể là tia phân giác của góc yOx' vì góc yOx' = 135 độ ko thể chia 2

\(5\widehat{xOy}=4\widehat{yOz}\Rightarrow\frac{\widehat{yOz}}{5}=\frac{\widehat{xOy}}{4}=\frac{\widehat{yOz}-\widehat{xOy}}{5-4}=\frac{10^0}{1}=10^0\)

\(\Rightarrow\widehat{yOz}=50^0;\widehat{xOy}=40^0\Rightarrow\widehat{yOz}+\widehat{xOy}=50^0+40^0=90^0\)

mà\(\widehat{yOz}+\widehat{xOy}=\widehat{xOz}\)(vì\(\widehat{xOy},\widehat{yOz}\)kề nhau) nên\(\widehat{xOz}=90^0\)=> Ox vuông góc Oz

a)Vì góc xOy và góc yOz là 2 góc kề bù

suy ra:xOy +yOz =180\(^0\)

          thay xOy =60\(^0\) có:

          60 \(^0\)+yOz =180\(^0\)

                 yOz =180\(^0\)-60\(^0\)

                 yOz =120\(^0\)

Vậy yOz=120\(^0\)

b)Vì Ot là tia phân giác của góc xOy

suy ra:xOt=tOy=xOy:2=60\(^0\):2=30\(^0\)(thay xOy=60\(^0\))

Vì Oy nằm giữa 2 tia Ot và Oz

suy ra:tOy+yOz=zOt

          thay tOy=30\(^0\);yOz=120\(^0\)

          30\(^0\)+120\(^0\) =zOt

              150\(^0\)    =zOt

Vậy zOt= 150\(^0\)

 Ta có: ∠yOz +  ∠xOy = 180\(^0\) ( hai góc kề bù )

                ∠yOz + 60\(^0\) = 180\(^0\) 

                          ∠yOz = 120\(^0\)  (1)

Ta có: ∠yOt = \(\dfrac{60^0}{2}\) = \(30^0\)  ( vì Ot là phân giác ∠xOy ) (2)

TỪ (1) VÀ (2) 

⇒  ∠yOz + ∠yOt = ∠zOt

         120\(^0\) + \(30^0\) = ∠zOt

                    \(150^0\)= ∠zOt

Vậy ∠zOt = \(150^0\)

Vì \(\dfrac{ }{xOy}\) và \(\dfrac{ }{yOz}\) là hai góc kề bù

⇒\(\dfrac{ }{xOy}\)+\(\dfrac{ }{yOz}\)=180⁰

60⁰+\(\dfrac{ }{yOz}\)=180⁰

\(\dfrac{ }{yOz}\)=180⁰-60⁰

\(\dfrac{ }{yOz}\)=120⁰

Ta có:  Ot là tia phân giác của góc \(\dfrac{ }{xOy}\)

⇒\(\dfrac{ }{xOt=yOt}\)\(=60^0:2=30^0\)

Trên cùng nửa mặt phẳng bờ chứa tia xz, có \(\dfrac{ }{xOt}\)<\(\dfrac{ }{xOz}\)(30⁰<180⁰)

⇒Tia Ot nằm giữa tia Ox và Oz

\(\dfrac{ }{xOt}\)+\(\dfrac{ }{zOt}\)=\(\dfrac{ }{xOz}\)

30⁰+\(\dfrac{ }{zOt}\)=180⁰

\(\dfrac{ }{zOt}\)=180⁰-30⁰

\(\dfrac{ }{zOt}\)=150⁰