đề của bn có sai ko đấy?

Ta có \(\widehat{aOc}+\widehat{bOc}=180^0\left(kề.bù\right)\)

\(\Rightarrow2\widehat{bOc}+\widehat{bOc}=180^0\\ \Rightarrow3\widehat{bOc}=180^0\\ \Rightarrow\widehat{bOc}=60^0\Rightarrow\widehat{aOc}=120^0\)

Vì \(ab\cap cd=O\) nên \(\left\{{}\begin{matrix}\widehat{bOc}=\widehat{aOd}=60^0\\\widehat{aOc}=\widehat{bOd}=120^0\end{matrix}\right.\left(các.cặp.góc.đối.đỉnh\right)\)

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\(\widehat{AOC}-\widehat{BOC}=50^o\)

\(\widehat{AOC}+\widehat{BOC}=\widehat{AOB}=180^o\)

suy ra \(\left\{{}\begin{matrix}\widehat{AOC}=\dfrac{180^o+50^o}{2}=115^o\\\widehat{BOC}=180^o-115^o=65^o\end{matrix}\right.\)

\(\widehat{AOD}=\widehat{BOC}=65^o\)

#)Giải :

#)Giải :

Vì \(\widehat{AOC}\)và \(\widehat{BOD}\)là hai góc đối đỉnh \(\Rightarrow\widehat{AOC}=\widehat{BOD}\left(=70^o\right)\)

Vì \(\widehat{AOC}\)và \(\widehat{BOC}\)là hai góc kề bù 

\(\Rightarrow\widehat{BOC}=180^o-\widehat{AOC}\)

               \(=180^o-70^o\)

               \(=110^o\)

\(\Rightarrow\widehat{BOC}=110^o\)

Vì \(\widehat{BOC}\)và \(\widehat{AOD}\)là hai góc đối đỉnh \(\Rightarrow\widehat{BOC}=\widehat{AOD}\left(=110^o\right)\)

                  #~Will~be~Pens~#

Theo đề bài biết :

\(\widehat{AOC}\)- \(\widehat{BOC}\)= 70^o

Ngoài ra còn biết :

\(\widehat{AOC}\)+ \(\widehat{BOC}\)= 180^o ( kề bù )

\(\rightarrow\)\(\widehat{AOC}\)= ( 70^o + 180^o ) : 2 = 125^o

\(\rightarrow\)\(\widehat{BOC}\)= 180^o - 125^o = 55^o

Có \(\widehat{AOD}\)+ \(\widehat{AOC}\)= 180^o ( kề bù )

\(\rightarrow\)\(\widehat{AOD}\)= 180^o - \(\widehat{AOC}\)= 180^o - 125^o = 55^o

Có \(\widehat{BOD}\)+ \(\widehat{BOC}\)= 180^o ( kề bù )

\(\rightarrow\)\(\widehat{BOD}\)= 180^o - \(\widehat{BOC}\)

180^o - 55^o = 125^o

Có : \(\widehat{AOC}=\widehat{BOD}\)( Hai góc đối đỉnh )

mà \(\widehat{AOC}=\widehat{BOD}\Rightarrow\widehat{BOD}=70^o\)

Có : \(\widehat{BOD}\)và \(\widehat{AOD}\)là hai góc kề bù 

\(\Rightarrow\widehat{BOD}+\widehat{AOD}=180^o\)

      \(70^o+\widehat{AOD}=180^o\Rightarrow\widehat{AOD}=180^o-70^o=110^o\)

Do \(\widehat{BOC}\)và \(\widehat{AOD}\)là hai góc đối đỉnh

\(\Rightarrow\widehat{BOC}=\widehat{AOD}\)

MÀ \(\widehat{AOD}=110^o\Rightarrow\widehat{BOC}=110^o\)

a) AOC^ + AOD^ = 180o (kề bù)

4*  AOD^ + AOD^ = 180o

5* AOD^ = 180o

AOD^ = 36o   =>  AOC^ = 144o

AOD^ = BOC^ = 36o (đđ)

AOC^ = BOD^ = 144o (đđ)

b)  COM^ = DON^ (đđ) ; BOM^ = AON^ (đđ);  COM^ = BOM^  => DON^ = AON^

c)  AON^ = AOD^ /2 = 36o/2 = 18o 

CON^ = AOC^ + AON^ = 144o + 18o = = 162o 

=> ON ko vuông góc OC. 

GIÚP MK VỚI

góc AOC=3*góc BOC

góc AOC+góc BOC=180 độ

=>góc AOC=3/4*180=135 độ; góc BOC=180-135=45 độ

góc AOD=góc BOC=45 độ

góc BOD=góc AOC=135 độ