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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 ...
cot C=2
=>\(tanC=\dfrac{1}{cotC}=\dfrac{1}{2}\)
\(1+tan^2C=\dfrac{1}{cos^2C}\)
=>\(cos^2C=1+\dfrac{1}{4}=\dfrac{5}{4}\)
=>\(cosC=\dfrac{2}{\sqrt{5}}\) hoặc \(cosC=-\dfrac{2}{\sqrt{5}}\)
TH1: \(cosC=\dfrac{2}{\sqrt{5}}\)
=>\(\dfrac{BC^2+AC^2-AB^2}{2\cdot BC\cdot AB}=\dfrac{2}{\sqrt{5}}\)
=>\(\dfrac{5+9-AB^2}{6\sqrt{5}}=\dfrac{2}{\sqrt{5}}\)
=>\(14-AB^2=12\)
=>AB^2=2
=>\(AB=\sqrt{2}\)
TH2: \(cosC=-\dfrac{2}{\sqrt{5}}\)
=>\(\dfrac{5+9-AB^2}{6\sqrt{5}}=-\dfrac{2}{\sqrt{5}}\)
=>\(14-AB^2=\dfrac{-2}{\sqrt{5}}\cdot6\sqrt{5}=-12\)
=>AB^2=26
=>\(AB=\sqrt{26}\)
Lời giải:
$p=\frac{AB+BC+AC}{2}=\frac{\sqrt{6}+\sqrt{3}+3}{2}$
Theo công thức Heron:
$S_{ABC}=\sqrt{p(p-AB)(p-BC)(p-AC)}=\frac{3+\sqrt{3}}{2}$
Bán kính đường tròn ngoại tiếp:
$R=\frac{AB.BC.AC}{4S}=\sqrt{2}$ (đvđd)
\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=\dfrac{5}{24}\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cosA=10a^2\)
+) Ta có: \(AB \bot AC \Rightarrow \overrightarrow {AB} \bot \overrightarrow {AC} \Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 0\)
+) \(\overrightarrow {AC} .\overrightarrow {BC} = \left| {\overrightarrow {AC} } \right|.\left| {\overline {BC} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right)\)
Ta có: \(BC = \sqrt {A{B^2} + A{C^2}} = \sqrt 2 \Leftrightarrow \sqrt {2A{C^2}} = \sqrt 2 \)\( \Rightarrow AC = 1\)
\( \Rightarrow \overrightarrow {AC} .\overrightarrow {BC} = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)
+) \(\overrightarrow {BA} .\overrightarrow {BC} = \left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|.\cos \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)
1.
Gọi M là trung điểm BC thì theo tính chất trọng tâm: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\Rightarrow x+y=\dfrac{2}{3}\)
2.
\(CH=\dfrac{1}{2}BC=\dfrac{a}{2}\)
\(T=\left|\text{ }\overrightarrow{CA}-\overrightarrow{HC}\right|=\left|\overrightarrow{CA}+\overrightarrow{CH}\right|\)
\(\Rightarrow T^2=CA^2+CH^2+2\overrightarrow{CA}.\overrightarrow{CH}=a^2+\left(\dfrac{a}{2}\right)^2+2.a.\dfrac{a}{2}.cos60^0=\dfrac{7a^2}{4}\)
\(\Rightarrow T=\dfrac{a\sqrt{7}}{2}\)
3.
\(10< x< 100\Rightarrow10< 3k< 100\)
\(\Rightarrow\dfrac{10}{3}< k< \dfrac{100}{3}\Rightarrow4\le k\le33\)
\(\Rightarrow\sum x=3\left(4+5+...+33\right)=1665\)
