Đề thi Học kì 1 Toán 10 năm 2023-2024 trường THPT Bình Chiểu (Tp.HCM)

$\cos B = \dfrac{c^2 + a^2 + b^2}{2ca}$

$\cos B = \dfrac{c^2 + a^2 - b^2}{ca}$

$\cos B = \dfrac{c^2 + a^2 - b^2}{2ca}$

$\cos B = \dfrac{c^2 + a^2 + b^2}{ca}$

$a^2 = b^2 + c^2 - 2bc \cos A$

$a^2 = b^2 + c^2 - bc \cos A$

$a^2 = b^2 + c^2 + 2bc \cos A$

$a^2 = b^2 + c^2 + bc \cos A$

$\dfrac{NP}{\sin N} = 2R$

$\dfrac{MP}{\sin N} = 2R$

$\dfrac{MN}{\sin N} = 2R$

$\dfrac{MP}{\sin N} = R$

$S = pr$

$S = \dfrac{a+b+c}{4r}$

$S = \dfrac{abc}{4r}$

$S = \dfrac{p}{r}$

$\overrightarrow{EF} . \overrightarrow{CD} = |\overrightarrow{EF}| |\overrightarrow{CD}| \cos \widehat{EF, CD}$

$\overrightarrow{EF} . \overrightarrow{CD} = \overrightarrow{EF} . \overrightarrow{CD} . \cos \widehat{EF, CD}$

$\overrightarrow{EF} . \overrightarrow{CD} = |\overrightarrow{EF}| . |\overrightarrow{CD}|$

$\overrightarrow{EF} = \overrightarrow{EF} . \overrightarrow{CD} \cos \widehat{EF, CD}$

$\widehat{A} = 50^\circ$

$\widehat{A} = 60^\circ$

$\widehat{A} = 60^\circ 7'$

$\widehat{A} = 50^\circ 7'$

$\overrightarrow{AC}$

$\overrightarrow{DB}$

$\overrightarrow{CB}$

$\overrightarrow{BC}$

$2\, cm$.

$3\sqrt{3}\, cm$.

$1\, cm$.

$3\, cm$.

$\widehat{(\overrightarrow{AC},\overrightarrow{CB})}=135^\circ$

$\widehat{(\overrightarrow{AC},\overrightarrow{CB})}=45^\circ$

$\widehat{(\overrightarrow{AC},\overrightarrow{CB})}=90^\circ$

$\widehat{(\overrightarrow{AC},\overrightarrow{CB})}=35^\circ$