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![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cwidehat%7BAOI%7D+%5Cwidehat%7B%20BOI%7D%3D%5Cwidehat%7BAOB%7D)
Suy ra
hay ![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cwidehat%7BAOI%7D%3D15%5E%7B0%7D)
\(\widehat{BOI}\) = \(\dfrac{1}{4}\) \(\widehat{AOB}\) ⇒ \(\dfrac{1}{4}\).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