Question

[Despacito]

cho tam giac ABC vuong tai A. Biet cos B = 0,8, hay tinh cac ti so luong giac cua goc C

Answer 1

[Ngo Tung Lam]

\(\cos B=\frac{AB}{BC}=0,8\)  mà  \(\sin C=\frac{AB}{BC}=\Rightarrow\sin C=0,8\)

Theo bài ra ta có :

\(\sin C^2+\cos C^2=\frac{AB}{BC}^2+\frac{AC}{BC}^2\)

\(=\frac{\left(AB^2+AC^2\right)}{BC^2}\)

\(=\frac{BC^2}{BC^2}\)

\(=1\)

\(\Rightarrow\cos C^2=1-\sin C^2=1-0,8^2=0,36\)

\(\Rightarrow\cos C=0,6\)hoặc \(\cos C=-0,6\)( loại vì C là một góc nhọn )

\(\Rightarrow\cos C=0,6\)

\(\Rightarrow\tan C=\frac{0,8}{0,6}=\frac{4}{3};\cot C=\frac{0,6}{0,8}=0,75\)

Vậy : \(\cos C=0,6\); \(\tan C=\frac{4}{3}\)và \(\cot C=0,75\)

Answer 2

[Despacito]

ta co : \(\sin^2B+\cos^2B=1\)

\(\Rightarrow\sin^2B=1-\cos^2B\)

\(\Rightarrow\sin^2B=1-\left(0,8\right)^2\)

\(\Rightarrow\sin^2B=1-0,64\)

\(\Rightarrow\sin^2B=0,36\)

\(\Rightarrow\sin B=0,6\)

ta co:   \(\tan B=\frac{\sin B}{\cos B}\)hay \(\tan B=\frac{0,6}{0,8}\)

\(\Rightarrow\tan B=0,75\)

ta co :  \(\cot B=\frac{\cos B}{\sin B}\)hay \(\cot B=\frac{0,8}{0,6}\)

\(\Rightarrow\cot B=\frac{4}{3}\)

+) \(B+C=90^0\)

\(\Rightarrow\sin B=\cos C=0,6\)

\(\Rightarrow\cos B=\sin C=0,8\)

\(\Rightarrow\tan B=\cot C=0,75\)

\(\Rightarrow\cot B=\tan C=\frac{4}{3}\)