Question

[Nguyễn Khánh Linh]

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 , \(a=14,b=18,c=20\)

Answer 1

[Nguyễn Việt Lâm]

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/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{11}{15}\Rightarrow A\approx42^050'\)

\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{17}{35}\Rightarrow B\approx60^056'\)

\(C=180^0-\left(A+B\right)=...\)

Answer 2

[Nguyễn Việt Lâm]

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/\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow sinB=\frac{bsinA}{a}\approx0,6\Rightarrow B\approx36^052'\)

\(\Rightarrow C=180^0-\left(A+B\right)=75^045'\)

\(\frac{a}{sinA}=\frac{c}{sinC}\Rightarrow c=\frac{a.sinC}{sinA}\approx21\)