Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Có \(\overrightarrow{BC}^2=\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{AC}^2+\overrightarrow{AB}^2-2\overrightarrow{AC}.\overrightarrow{AB}\)
Suy ra: \(\overrightarrow{AC}.\overrightarrow{AB}=\dfrac{\overrightarrow{AC^2}+\overrightarrow{AB}^2-\overrightarrow{BC}^2}{2}=\dfrac{8^2+6^2-11^2}{2}=-\dfrac{21}{2}\).
Do \(\overrightarrow{AC}.\overrightarrow{AB}< 0\) nên \(cos\widehat{BAC}< 0\) suy ra góc A là góc tù.
b) Từ câu a suy ra: \(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=-\dfrac{21}{2.6.8}=-\dfrac{7}{32}\).
Do N là trung điểm của AC nên \(AN=AC:2=8:2=4cm\).
\(\overrightarrow{AM}.\overrightarrow{AN}=AM.AN.cos\left(\overrightarrow{AM},\overrightarrow{AN}\right)\)
\(=2.4.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=2.4.\dfrac{-7}{32}=-\dfrac{7}{4}\).
A B C I M H J K
a. ta có \(BI=\frac{1}{4}BA=\frac{3}{4}\)
Dễ thấy hai tam giác \(\Delta ABM~\Delta CBI\Rightarrow\frac{MB}{IB}=\frac{AB}{BC}\Rightarrow MB=\frac{3}{4}.\frac{3}{4}=\frac{9}{16}\)
vậy \(\frac{BM}{BC}=\frac{9}{64}\).
b.Xét tam giác AJB ta áp dụng địh lý menelaus có
\(\frac{AC}{CJ}.\frac{JK}{KB}.\frac{BI}{IA}=1\Rightarrow\frac{JK}{KB}=\frac{3}{2}\Rightarrow\frac{BK}{KJ}=\frac{2}{3}\)
a) Có
\(\overrightarrow{BC}^2=\left(\overrightarrow{BA}+\overrightarrow{AC}\right)^2=\overrightarrow{BA}^2+\overrightarrow{AC}^2+2\overrightarrow{BA}.\overrightarrow{AC}\)
\(=\overrightarrow{BA}^2+\overrightarrow{AC}^2-2\overrightarrow{AB}.\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{\overrightarrow{BA}^2+\overrightarrow{AC}^2-\overrightarrow{BC^2}}{2}=\dfrac{5^2+8^2-7^2}{2}=20\).
\(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=\dfrac{20}{5.8}=\dfrac{1}{2}\).
Vì vậy \(\widehat{BAC}=60^o\).
b) Tương tự:
\(\overrightarrow{CA}.\overrightarrow{CB}=\dfrac{CA^2+CB^2-AB^2}{2}=\dfrac{7^2+8^2-5^2}{2}=44\).
Lời giải:
\(|\overrightarrow{AB}|=BC\cos B=2.\cos 60^0=1\) (cm)
\(|\overrightarrow{AC}|=BC\sin B=2.\sin 60^0=\sqrt{3}\) (cm)
------------------
Do tam giác $ABC$ vuông tại $A$ nên $\overrightarrow{AB}\perp \overrightarrow{AC}\Rightarrow \overrightarrow{AB}.\overrightarrow{AC}=0$. Do đó:
\(|\overrightarrow{AB}+\overrightarrow{AC}|^2=(\overrightarrow{AB}+\overrightarrow{AC})^2=AB^2+AC^2+2\overrightarrow{AB}.\overrightarrow{AC}\)
\(=BC^2+0=BC^2=4\) (cm)
$\Rightarrow |\overrightarrow{AB}+\overrightarrow{AC}|=2$ (cm)
Tương tự:
\(|\overrightarrow{AB}-\overrightarrow{AC}|^2=AB^2+AC^2-2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2=BC^2=4\)
$\Rightarrow |\overrightarrow{AB}-\overrightarrow{AC}|=2$ (cm)
Lời giải:
\(|\overrightarrow{AB}|=BC\cos B=2.\cos 60^0=1\) (cm)
\(|\overrightarrow{AC}|=BC\sin B=2.\sin 60^0=\sqrt{3}\) (cm)
------------------
Do tam giác $ABC$ vuông tại $A$ nên $\overrightarrow{AB}\perp \overrightarrow{AC}\Rightarrow \overrightarrow{AB}.\overrightarrow{AC}=0$. Do đó:
\(|\overrightarrow{AB}+\overrightarrow{AC}|^2=(\overrightarrow{AB}+\overrightarrow{AC})^2=AB^2+AC^2+2\overrightarrow{AB}.\overrightarrow{AC}\)
\(=BC^2+0=BC^2=4\) (cm)
$\Rightarrow |\overrightarrow{AB}+\overrightarrow{AC}|=2$ (cm)
Tương tự:
\(|\overrightarrow{AB}-\overrightarrow{AC}|^2=AB^2+AC^2-2\overrightarrow{AB}.\overrightarrow{AC}=AB^2+AC^2=BC^2=4\)
$\Rightarrow |\overrightarrow{AB}-\overrightarrow{AC}|=2$ (cm)
Đặt AB = c ; AC = b ; BC = a .
Ta có : \(b+c=13\) ; \(r=\dfrac{S}{p}=\sqrt{3}\) ( p \(=\dfrac{a+b+c}{2}\) )
Có : \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\) nên : \(r=\sqrt{\dfrac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}}=\sqrt{3}\)
\(\Rightarrow\left(p-a\right)\left(p-b\right)\left(p-c\right)=3p\)
\(\Leftrightarrow\left(\dfrac{-a+b+c}{2}\right)\left(\dfrac{-b+a+c}{2}\right)\left(\dfrac{-c+a+b}{2}\right)=\dfrac{3\left(a+b+c\right)}{2}\)
\(\Leftrightarrow\left(-a+b+c\right)\left(-b+a+c\right)\left(-c+a+b\right)=12\left(a+b+c\right)\)
\(\Leftrightarrow\left(-a+13\right)\left(-b+a+c\right)\left(-c+a+b\right)=12\left(13+a\right)\)
\(\Leftrightarrow\left(-a+13\right)\left[a^2-\left(b-c\right)^2\right]=12\left(13+a\right)\) (2)
Có : \(\dfrac{b^2+c^2-a^2}{2bc}=cosA=cos60^o=\dfrac{1}{2}\) \(\Rightarrow b^2+c^2-a^2=bc\) \(\Leftrightarrow a^2=b^2+c^2-bc\) (1)
Mặt khác : \(b+c=13\Leftrightarrow b^2+c^2-bc+3bc=169\Leftrightarrow a^2=169-3bc\)
Từ (1) ; (2) suy ra : \(\left(-a+13\right)bc=12\left(13+a\right)\)
\(\Leftrightarrow\left(-a+13\right)\left(169-a^2\right)=36\left(13+a\right)\)
\(\Leftrightarrow\left(13-a\right)^2\left(13+a\right)=36\left(13+a\right)\)
\(\Leftrightarrow\left(13-a\right)^2=36\) \(\Leftrightarrow\left[{}\begin{matrix}13-a=6\\13-a=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=7\\a=19>13=b+c\left(L\right)\end{matrix}\right.\)
Vậy ...