\(\frac{3}{4}=tanB=\frac{AC}{AB}\Rightarrow AC=\frac{3}{4}AB\)

\(BC^2=AB^2+AC^2\)(định lí Pythagore) 

\(=AB^2+\frac{9}{16}AB^2=\frac{25}{16}AB^2\)

\(\Rightarrow AB^2=10^2.\frac{16}{25}\Rightarrow AB=8\left(cm\right)\)

\(\Rightarrow AC=\frac{3}{4}AB=6\left(cm\right)\)

\(AC=\sqrt{BC^2-AB^2}=10\sqrt{3}\left(cm\right)\left(pytago\right)\\ \sin B=\dfrac{AC}{BC}=\dfrac{\sqrt{3}}{2}=\sin60^0\Rightarrow\widehat{B}=60^0\\ \widehat{C}=90^0-\widehat{B}=30^0\\ 2,\sin B\cdot\tan B=\dfrac{AC}{AB}\cdot\dfrac{AC}{BC}=\dfrac{AC^2}{AB\cdot BC}=\dfrac{HC\cdot BC}{AB\cdot BC}=\dfrac{HC}{AB}\\ 3,\dfrac{CI}{IB}=\dfrac{AC}{AB}=\sqrt{3}\Leftrightarrow CI=\sqrt{3}IB\\ CI+IB=BC=20\\ \Rightarrow\left(\sqrt{3}+1\right)IB=20\Leftrightarrow IB=\dfrac{20}{\sqrt{3}+1}=10\sqrt{3}-10\left(cm\right)\\ HB=\dfrac{AB^2}{BC}=5\left(cm\right)\left(HTL\right)\\ IH=IB-HB=10\sqrt{3}-15\left(cm\right)\)

\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{4}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{3}{4}\\ \Rightarrow\cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{4}{3}\)

Trong tam giác vuông ABC , ta có:

Áp dụng định lí Py-ta-go trong tam giác vuông ABC, ta có:

\(AC=\sqrt{AB^2-BC^2}=\sqrt{\left(\dfrac{3}{2}BC\right)^2-BC^2}=\dfrac{BC\sqrt{5}}{2}\)

Ta có:

 \(\tan B=\dfrac{AC}{BC}=\dfrac{\dfrac{BC\sqrt{5}}{2}}{BC}=\dfrac{\sqrt{5}}{2}\)

Chọn đáp án D

Trong tam giác vuông ABC $\left(\hat{C}={90}^0\right)$, ta có:

$\sin A=\frac{BC}{AB}=\frac{2}{3}\Rightarrow AB=\frac{3}{2}BC$

Áp dụng định lí Py-ta-go trong tam giác vuông ABC, ta có:

$\begin{array}{c}AC=\sqrt{A{B}^2-B{C}^2}=\sqrt{{\left(\frac{3}{2}BC\right)}^2-B{C}^2}\\ AC=\sqrt{\frac{5}{4}B{C}^2}=\frac{BC\sqrt{5}}{2}\end{array}$

Ta có: $\tan B=\frac{AC}{BC}=\frac{BC\frac{\sqrt{5}}{2}}{BC}=\frac{\sqrt{5}}{2}$

Chọn đáp án D