Hình 27 cho biết tia OI nằm giữa hai tia OA, OB, \(\widehat{AOB}=60^0;\widehat{BOI}=\dfrac{1}{4}\widehat{AOB}\)
Tính \(\widehat{BOI},\widehat{AOI}\) ?
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Tui cx ms hok thôi
*Theo đề bài ,ta có \(\widehat{BOI}=\frac{1}{4}.\widehat{AOB}\)
Hay\(\widehat{BOI}=\frac{1}{4}.60\)
\(\Rightarrow\widehat{BOI}=15^0\)
*Vì tia OI nằm giữa hai tia OA,OB
Nên\(\widehat{AOI}+\widehat{IOB}=\widehat{AOB}\)
Hay\(\widehat{AOI}+15^o=60^0\)
\(\Rightarrow\widehat{AOI}=60^o-15^o=45^0\)
Vậy \(\widehat{AOI}=45^o\)
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
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: \(\widehat{BOI}=15^0\); \(\widehat{AOI}=45^0\)
\(\widehat{BOI}=\frac{1}{4}\widehat{AOB}\Rightarrow\widehat{BOI}=\frac{1}{4}.60^o\Rightarrow\widehat{BOI}=15^o\)
Vì tia OI nằm giữa hai tia OA và OB
\(\Rightarrow\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)
Thay \(\widehat{BOI}=15^o,\widehat{AOB}=60^o\), ta có:
\(\widehat{AOI}+15^o=60^o\)
\(\Rightarrow\widehat{AOI}=60^o-15^o\)
\(\Rightarrow\widehat{AOI}=45^o\)
Vậy \(\widehat{BOI}=15^o,\widehat{AOI}=45^o\)
\(\widehat{BOI}=\frac{1}{4}\widehat{AOB}=>\widehat{BOI}=\frac{1}{4}\cdot60^O\)\(=>\widehat{BOI}=15^O\)
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 \(\widehat{BOI}=15^o;\widehat{AOI}=45^o\)
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 : \(\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)
\(\Rightarrow\widehat{AOI}+15^o=60^o\)\(\Rightarrow\widehat{AOI}=45^o\)
\(\widehat{BOI}=60':4=15'\)
\(\widehat{AOI}=60'-15'=45'\)
\('\) : độ
Giải
Do tia OI nằm giũa hai tia OA, OB nên
Suy ra hay
\(\widehat{BOI}\) = \(\dfrac{1}{4}\) \(\widehat{AOB}\) ⇒ \(\dfrac{1}{4}\).60o = 15o
Vì tia OI nằm giữa 2 tia OA; OB nên:
\(\widehat{AOI}\)+ \(\widehat{IOB}\)= \(\widehat{AOB}\)
⇒ \(\widehat{AOI}\)= \(\widehat{AOB} - \widehat{IOB}\) = 60o - 15o = 45o
Vậy: \(\widehat{AOI}\) = 45o