cho tam giác ABC có AB=25cm,AC=24cm,BC=36cm.tính số đo 3 góc 02/08/2021 Bởi Gianna cho tam giác ABC có AB=25cm,AC=24cm,BC=36cm.tính số đo 3 góc
Áp dụng định lý $\cos$ ta được: $+) \quad BC^2 = AB^2 + AC^2 – 2AB.AC\cos\widehat{BAC}$ $\Rightarrow \cos\widehat{BAC} = \dfrac{AB^2+ AC^2 – BC^2}{2AB.AC} = \dfrac{25^2 + 24^2 – 36^2}{2.25.24} = -\dfrac{19}{240}$ $\Rightarrow \widehat{BAC} = \arccos\left( -\dfrac{19}{240}\right) \approx 94,54^o$ $+) \quad AB^2 = AC^2 + BC^2 – 2AC.BC\cos\widehat{ACB}$ $\Rightarrow \cos\widehat{ACB} = \dfrac{AC^2 + BC^2 – AB^2}{2AC.BC} = \dfrac{24^2 + 36^2 – 25^2}{2.24.36} = \dfrac{1247}{1728}$ $\Rightarrow \widehat{ACB} = \arccos\dfrac{1247}{1728} \approx 43,81^o$ $\Rightarrow \widehat{ABC} = 180^o – (\widehat{BAC} + \widehat{ACB}) = 180^o – (94,54 + 43,81) = 41,65^o$ Bình luận
Áp dụng định lý $\cos$ ta được:
$+) \quad BC^2 = AB^2 + AC^2 – 2AB.AC\cos\widehat{BAC}$
$\Rightarrow \cos\widehat{BAC} = \dfrac{AB^2+ AC^2 – BC^2}{2AB.AC} = \dfrac{25^2 + 24^2 – 36^2}{2.25.24} = -\dfrac{19}{240}$
$\Rightarrow \widehat{BAC} = \arccos\left( -\dfrac{19}{240}\right) \approx 94,54^o$
$+) \quad AB^2 = AC^2 + BC^2 – 2AC.BC\cos\widehat{ACB}$
$\Rightarrow \cos\widehat{ACB} = \dfrac{AC^2 + BC^2 – AB^2}{2AC.BC} = \dfrac{24^2 + 36^2 – 25^2}{2.24.36} = \dfrac{1247}{1728}$
$\Rightarrow \widehat{ACB} = \arccos\dfrac{1247}{1728} \approx 43,81^o$
$\Rightarrow \widehat{ABC} = 180^o – (\widehat{BAC} + \widehat{ACB}) = 180^o – (94,54 + 43,81) = 41,65^o$