\(a,BC=\sqrt{AB^2+AC^2}=15\left(cm\right)\left(pytago\right)\)

Áp dụng HTL:

\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\\AH^2=BH\cdot CH\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=5,4\left(cm\right)\\CH=\dfrac{AC^2}{BC}=9,6\left(cm\right)\\AH=\sqrt{5,4\cdot9,6}=51,84\left(cm\right)\end{matrix}\right.\)

\(b,\sin B=\cos C=\dfrac{AC}{BC}=\dfrac{4}{5}\\ \cos B=\sin C=\dfrac{AB}{BC}=\dfrac{3}{5}\\ \tan B=\cot C=\dfrac{AC}{AB}=\dfrac{4}{3}\\ \cot B=\tan C=\dfrac{AB}{AC}=\dfrac{3}{4}\)

a: Xét ΔMNE vuông tại M có 

\(MN^2+ME^2=NE^2\)

hay ME=4(cm)

GF=căn 0,9^2+1,2^2=1,5cm

sin G=cos F=EF/GF=3/5

cosG=sin F=EG/GF=4/5

tan G=cot F=3/5:4/5=3/4

cot G=tan F=4/3

Xét ΔABC vuông tại A có 

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

hay AC=8(cm)

Xét ΔABC vuông tại A có 

\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{8}{10}=\dfrac{4}{5}\)

\(\cos\widehat{B}=\dfrac{3}{5}\)

\(\tan\widehat{B}=\dfrac{4}{3}\)

\(\cot\widehat{B}=\dfrac{3}{4}\)

Bài 2:

Sửa đề: \(\sin\alpha=\dfrac{3}{5}\)

Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\)

\(\Leftrightarrow\cos\alpha=\dfrac{4}{5}\)

Ta có: \(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\)

\(=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{5}\cdot\dfrac{5}{4}=\dfrac{3}{4}\)

Ta có: \(\cot\alpha=\dfrac{1}{\tan\alpha}\)

\(=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

tan B=3/4

=>AC/AB=3/4

=>AC=4,5

BC=căn AB^2+AC^2=7,5

sin C=AB/BC=6/7,5=4/5

cos C=AC/BC=3/5

tan C=4/3

cot C=3/4