\(\widehat{BOI}=60':4=15'\)

\(\widehat{AOI}=60'-15'=45'\)

\('\) : độ

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: \(\widehat{BOI}=15^0\); \(\widehat{AOI}=45^0\)

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

\(\widehat{BOI}=\frac{1}{4}\widehat{AOB}=>\widehat{BOI}=\frac{1}{4}\cdot60^O\)\(=>\widehat{BOI}=15^O\)

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 \(\widehat{BOI}=15^o;\widehat{AOI}=45^o\)

Hình vẽ đây các bạn

góc BOI=1/5*60=12 độ

góc AOI=60-12=48 độ

\(\widehat{BOI}=\dfrac{1}{5}\cdot60^0=12^0\)

\(\widehat{AOI}=60^0-12^0=48^0\)

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 : \(\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)

\(\Rightarrow\widehat{AOI}+15^o=60^o\)\(\Rightarrow\widehat{AOI}=45^o\)

\(\widehat{AOI}=180^0-120^0=60^0\)

\(\widehat{KOB}=180^0-45^0=135^0\)

\(\widehat{BOA}=60^0+45^0=105^0\)