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Tia Oc nằm giữa 2 tia OA,OB nên

\(\widehat{AoC}\)\(+\)\(\widehat{CoB}\)\(=\)\(\widehat{AoB}\) \(\left(1\right)\)

=>\(\widehat{Aoc}+\widehat{CoB}\)\(=90^0\)

Theo đề ta có \(\frac{1}{4}AoC=\frac{1}{5}CoB\left(2\right)\)

Từ \(\left(2\right)\) \(\Rightarrow AoC=\frac{4}{5}CoB\)

Thay \(\frac{4}{5}CoB+CoB=90^0\)

\(=\frac{9}{5}CoB=90^0\)

\(CoB=90^0\div\frac{9}{5}=50^0\)

tia OC nằm giữa hai tia OA và OB

nên \(\widehat{COA}+\widehat{COB}=\widehat{AOB}=90^0\)

mà \(\widehat{AOC}-2\cdot\widehat{COB}=30^0\)

nên \(\widehat{AOC}+\widehat{COB}-\widehat{AOC}+2\cdot\widehat{COB}=90^0-30^0=60^0\)

=>\(3\cdot\widehat{COB}=60^0\)

=>\(\widehat{COB}=20^0\)

=>\(\widehat{AOC}=2\cdot20^0+30^0=70^0\)

AOC=60 độ

COB= 30 độ​

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1) Trên cùng nửa mặt phẳng bờ OA, ta có \(\widehat{AOB}< \widehat{AOC}\)nên OB nằm giữa OA, OC, suy ra \(\widehat{AOB}+\widehat{BOC}=\widehat{AOC}\)

OD là phân giác \(\widehat{AOB}\)nên AD nằm giữa OA, OB, suy ra \(\widehat{AOD}+\widehat{DOB}=\widehat{AOB}\). Ngoài ra, \(\widehat{AOD}=\widehat{DOB}< \widehat{AOB}\)

\(\widehat{AOD}< \widehat{AOB};\widehat{AOB}< \widehat{AOC}\Rightarrow\widehat{AOD}< \widehat{AOC}\).

Trên cùng nửa mặt phẳng bờ OA, ta có \(\widehat{AOD}< \widehat{AOC}\)nên OD nằm giữa OA,OC, suy ra \(\widehat{AOD}+\widehat{DOC}=\widehat{AOC}\)

\(\Leftrightarrow\widehat{AOD}+\widehat{DOC}=\widehat{AOB}+\widehat{BOC}\Leftrightarrow\widehat{AOD}+\widehat{DOC}=\widehat{AOD}+\widehat{DOB}+\widehat{BOC}\)

\(\Leftrightarrow\widehat{DOC}=\widehat{DOB}+\widehat{BOC}\Leftrightarrow\) OB nằm giữa OD, OC

2) \(\frac{\widehat{COB}+\widehat{COA}}{2}=\frac{\widehat{COB}+\widehat{AOD}+\widehat{DOB}+\widehat{BOC}}{2}=\frac{2\left(\widehat{COB}+\widehat{DOB}\right)}{2}=\widehat{COD}\)

CON HEO