cho góc aob =60 độ ,oi nằm giữa 2 tia oa và ob tính aoi bt boi=1/4aob
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\widehat{BOI}=\dfrac{1}{5}\cdot60^0=12^0\)
\(\widehat{AOI}=60^0-12^0=48^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\widehat{BOI}=\dfrac{1}{4}\cdot\widehat{AOB}\)(gt)
\(\Leftrightarrow\widehat{BOI}=\dfrac{1}{4}\cdot60^0\)
hay \(\widehat{BOI}=15^0\)
Ta có: tia OI nằm giữa hai tia OA và OB(gt)
nên \(\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)
\(\Leftrightarrow\widehat{AOI}=\widehat{AOB}-\widehat{BOI}=60^0-15^0\)
hay \(\widehat{AOI}=45^0\)
Vậy: \(\widehat{BOI}=15^0\); \(\widehat{AOI}=45^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\widehat{AOB}\)= 600 và \(\widehat{BOI}\)= \(\frac{1}{4}\)\(\widehat{AOB}\)
\(=>\widehat{BOI}\)= \(\frac{1}{4}.60\)
\(=>\widehat{BOI}\)= 150
Ta lại có: \(\widehat{BOI}\)\(+\widehat{AOI}\)= \(\widehat{AOB}\)
Hay: 150 + \(\widehat{AOI}\)= 600
=> \(\widehat{AOI}\)= 450
Vẽ hình hơi xấu, thông cảm nhé ^^
Hk tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\widehat{BOI}=\frac{1}{4}\widehat{AOB}=>\widehat{BOI}=\frac{1}{4}\cdot60^O\)\(=>\widehat{BOI}=15^O\)
Vì tia OI nằm giữa hai tia OA và OB
\(=>\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)
Thay \(\widehat{BOI}=15^o,\widehat{AOB}=60^o\),ta có:
\(\widehat{AOI}+15^o=60^o\)
\(=>\widehat{AOI}=60^o-15^o\)
\(=>\widehat{AOI}=45^o\)
Vậy \(\widehat{BOI}=15^o;\widehat{AOI}=45^o\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\widehat{BOI}=\frac{1}{4}\widehat{AOB}\)
\(\Rightarrow\widehat{BOI}=\frac{1}{4}.60^o=15^o\)
OI nằm giữa OA và OB nên ta có : \(\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)
\(\Rightarrow\widehat{AOI}+15^o=60^o\)\(\Rightarrow\widehat{AOI}=45^o\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\widehat{BOI}=60':4=15'\)
\(\widehat{AOI}=60'-15'=45'\)
\('\) : độ