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Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)
\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:
a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)
\(=\cos150^o+\sin30^o+\tan60^o\)
\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)
\(=\frac{\sqrt{3}+1}{2}\)
b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)
\(=\sin90^o+\cos30^o+\cos0^o\)
\(=1+\frac{\sqrt{3}}{2}\)
\(=\frac{2+\sqrt{3}}{2}\)
A B C
a) \(\overrightarrow{AB}.\overrightarrow{AC}=0\) do \(AB\perp AC\).
b)
\(BC=\sqrt{AB^2+AC^2}=\sqrt{a^2+a^2}=\sqrt{2}a\).
\(\overrightarrow{BA}.\overrightarrow{BC}=BA.BC.cos\left(\overrightarrow{BA},\overrightarrow{BC}\right)=a.\sqrt{2}a.cos45^o=a^2\).
c) \(\overrightarrow{AB}.\overrightarrow{BC}=-\overrightarrow{BA}.\overrightarrow{BC}=-a^2\).
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)
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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)
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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)
Gọi M là trung điểm của BC
Xét ΔABC có AM là đường trung tuyến
nên \(\overrightarrow{AB}+\overrightarrow{AC}=2\cdot\overrightarrow{AM}\)
\(\Leftrightarrow\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=2\cdot\dfrac{a\sqrt{3}}{2}=a\sqrt{3}\)
\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{AB}\right|=CB=a\)
\(BC=AD=\sqrt{AC^2-AB^2}=2a\)
a/ \(T=\left|3\overrightarrow{AB}-4\overrightarrow{BC}\right|\Rightarrow T^2=9AB^2+16BC^2-24\overrightarrow{AB}.\overrightarrow{BC}\)
\(=9a^2+64a^2=73a^2\Rightarrow T=a\sqrt{73}\)
b/ \(T^2=4AB^2+9BC^2+12.\overrightarrow{BA}.\overrightarrow{BC}=4AB^2+9BC^2=40a^2\)
\(\Rightarrow T=2a\sqrt{10}\)
c/ \(T=\left|\overrightarrow{AD}+3\overrightarrow{BC}\right|=\left|\overrightarrow{AD}+3\overrightarrow{AD}\right|=\left|4\overrightarrow{AD}\right|=4AD=8a\)
d/ \(T=\left|2\overrightarrow{DC}-3\overrightarrow{DC}\right|=\left|-\overrightarrow{DC}\right|=CD=AB=a\)
ta có ΔABC vuông tại A với AB=2a \(BC=a\sqrt{5}\)
=>AC=\(\sqrt{BC-AB}=\sqrt{5a^2-4a^2}=a\)
vẽ AK=2AB=>AK=4a; AH=5AC=>AH=5a
=>\(\overrightarrow{AK}=2\overrightarrow{AB};\overrightarrow{AH}=5\overrightarrow{AC}\)
khi đó \(\left|\overrightarrow{2AB}-5\overrightarrow{AC}\right|=\left|\overrightarrow{AK}-\overrightarrow{AH}\right|=\left|\overrightarrow{HK}\right|=HK\)
xét tam giác vuông AHK có AK=4a ;AH=5a
=>HK=\(\sqrt{AK^2+AH^2}=\sqrt{16a^2+25a^2}=a\sqrt{41}\)