^O1=^O3, ^O2=^O4 ( đối đỉnh)
a, O1+O2=180độ -> 1/3O2+O2=180 độ -> O2=135o-> O1=45o
b, O2-O1=50, O1+O2=180o -> O2=(180o+50o)/2=115o -> O1=115-50=65o
c, O1+O3=1/2(O2+O4)-> 2O1=O2-> 2O1+O1=180o-> O1=60o-> O2=120o
Giải:
a) \(\widehat{O}_1=\widehat{O}_3,\widehat{O}_2=\widehat{O}_4\) (đối đỉnh)
\(\Rightarrow\widehat{O}_1+\widehat{O}_2=180^o\)
\(\Rightarrow\dfrac{1}{3}\widehat{O}_2+\widehat{O}_2=180^o\)
\(\Rightarrow\widehat{O}_2=135^o\)
\(\Rightarrow\widehat{O}_1=45^o\)
\(\Rightarrow\widehat{O}_3=45^o\)
\(\Rightarrow\widehat{O}_4=135^o\)
b) \(\widehat{O}_2-\widehat{O}_1=50^o,\widehat{O}_1+\widehat{O}_2=180^o\)
\(\Rightarrow\widehat{O}_2=\left(180^o+50^o\right):2=115^o\)
\(\Rightarrow\widehat{O}_1=115^o-50^o=65^o\)
\(\Rightarrow\widehat{O}_3=65^o\)
\(\Rightarrow\widehat{O}_4=115^o\)
c) \(\widehat{O}_1+\widehat{O}_3=\dfrac{1}{2}\left(\widehat{O}_2+\widehat{O}_4\right)\)
\(\Rightarrow2\widehat{O}_1=\widehat{O}_2\)
\(\Rightarrow2\widehat{O}_1+\widehat{O}_1=180^o\)
\(\Rightarrow\widehat{O}_1=60^o\)
\(\Rightarrow\widehat{O}_2=120^o\)
\(\Rightarrow\widehat{O}_3=60^o\)
\(\Rightarrow\widehat{O}_4=120^o\)
