Bài 1:

a) Ta có:

\(tanB=\dfrac{AC}{AB}\Rightarrow\dfrac{AC}{AB}=\dfrac{5}{2}\)

\(\Rightarrow AC=\dfrac{AB\cdot5}{2}=\dfrac{6\cdot5}{2}=15\)  

b) Áp dụng Py-ta-go ta có: 

\(BC^2=AB^2+AC^2=6^2+15^2=261\)

\(\Rightarrow BC=\sqrt{261}=3\sqrt{29}\)

Bài 2: 

\(\left\{{}\begin{matrix}sinM=sin40^o\approx0,64\Rightarrow cosN\approx0,64\\cosM=cos40^o\approx0,77\Rightarrow sinN\approx0,77\\tanM=tan40^o\approx0,84\Rightarrow cotN\approx0,84\\cotM=cot40^o\approx1,19\Rightarrow tanN\approx1,19\end{matrix}\right.\)

Ta có:

\(cosB=\dfrac{AB}{BC}\Rightarrow AB=BC.cosB=10.0,8=8\left(cm\right)\)

\(AC=\sqrt{BC^2-AB^2}=6\left(cm\right)\)

b.

\(sinC=\dfrac{AB}{BC}=\dfrac{8}{10}=0,8\)

\(cosC=\dfrac{AC}{BC}=\dfrac{6}{10}=0,6\)

\(tanC=\dfrac{AB}{AC}=\dfrac{8}{6}=\dfrac{4}{3}\)

\(cotC=\dfrac{AC}{AB}=\dfrac{3}{4}\)

a: \(\widehat{B}=60^0\)

AB=8cm

\(AC=4\sqrt{3}\left(cm\right)\)

k biết mới học lớp 5 thôi

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow AC^2=3.6^2-2.1^2=8.55\)

hay \(AC=\dfrac{3\sqrt{95}}{10}\left(cm\right)\)

Xét ΔABC vuông tại B có 

\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{3\sqrt{95}}{10}:\dfrac{36}{10}=\dfrac{\sqrt{95}}{12}\)

\(\cos\widehat{B}=\dfrac{AB}{BC}=\dfrac{2.1}{3.6}=\dfrac{7}{12}\)

\(\tan\widehat{B}=\dfrac{AC}{AB}=\dfrac{3\sqrt{95}}{10}:\dfrac{21}{10}=\dfrac{\sqrt{95}}{7}\)

\(\cot\widehat{B}=\dfrac{AB}{AC}=\dfrac{7\sqrt{95}}{95}\)