Giải tam giác biết
1 , \(a=2\sqrt{3},b=2\sqrt{2},c=\sqrt{6}-\sqrt{2}\)
2 , \(a=109,\widehat{B}=33^o24',\widehat{C}=66^o59'\)
3 , \(a=20,b=13,\widehat{A}=67^o23'\)
4 , \(b=4,5,\widehat{A}=30^o,\widehat{C}=75^o\)
5 , \(b=14,c=10,\widehat{A}=145^o\)
6 ,...
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Giải tam giác biết
1 , \(a=2\sqrt{3},b=2\sqrt{2},c=\sqrt{6}-\sqrt{2}\)
2 , \(a=109,\widehat{B}=33^o24',\widehat{C}=66^o59'\)
3 , \(a=20,b=13,\widehat{A}=67^o23'\)
4 , \(b=4,5,\widehat{A}=30^o,\widehat{C}=75^o\)
5 , \(b=14,c=10,\widehat{A}=145^o\)
6 , \(a=14,b=18,c=20\)
d/ \(B=180^0-\left(A+C\right)=75^0\)
\(\Rightarrow b=c=4,5\)
\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow a=\frac{b.sinA}{sinB}=\frac{9}{4}\left(\sqrt{6}-\sqrt{2}\right)\)
e/ \(cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow a=\sqrt{b^2+c^2-2bc.cosA}\approx23\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{433}{460}\Rightarrow B\approx19^043'\)
\(\Rightarrow C=180^0-\left(A+B\right)=...\)
f/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{11}{15}\Rightarrow A\approx42^050'\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{17}{35}\Rightarrow B\approx60^056'\)
\(C=180^0-\left(A+B\right)=...\)
a/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=-\frac{1}{2}\Rightarrow A=120^0\)
\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)
\(C=180^0-\left(A+B\right)=15^0\)
b/\(A=180^0-\left(B+C\right)=79^037'\)
\(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\Rightarrow\left\{{}\begin{matrix}b=\frac{sinB}{sinA}.a\approx61\\c=\frac{sinC}{sinA}.a\approx102\end{matrix}\right.\)
c/\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow sinB=\frac{bsinA}{a}\approx0,6\Rightarrow B\approx36^052'\)
\(\Rightarrow C=180^0-\left(A+B\right)=75^045'\)
\(\frac{a}{sinA}=\frac{c}{sinC}\Rightarrow c=\frac{a.sinC}{sinA}\approx21\)