TA CÓ: OI LÀ TIA NẰM GIỮA OA; OB

\(\Rightarrow\widehat{AOI}+\widehat{IOB}=\widehat{AOB}\)

MÀ: \(\widehat{AOI}=\frac{1}{4}\widehat{IOB}\)

THAY SỐ: \(\frac{1}{4}\widehat{IOB}+\widehat{IOB}=70^0\)

                 \(\Rightarrow\left(\frac{1}{4}+1\right)\widehat{IOB}=70^0\)

                 \(\frac{5}{4}\widehat{IOB}=70^0\)

                      \(\widehat{IOB}=70^0:\frac{5}{4}\)

\(\Rightarrow\widehat{IOB}=56^0\)

MÀ \(\widehat{AOI}=\frac{1}{4}\widehat{IOB}\)

THAY SỐ: \(\widehat{AOI}=\frac{1}{4}.56^0=14^0\)

\(\Rightarrow\widehat{AOI}=14^0\)

CHÚC BN HỌC TỐT!!!!!

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

Đáp án là B

KOB = 135

AOI = 60

A0B =105

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

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

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