Đáp án:
$\begin{array}{l}
1)\\
+ A{H^2} = BH.CH = 3,2.1,8 = 5,76\\
\Rightarrow AH = \sqrt {5,76} = 2,4\left( {cm} \right)\\
+ BC = CH + BH = 3,2 + 1,8 = 5\left( {cm} \right)\\
+ A{B^2} = BH.BC = 3,2.5 = 16\\
\Rightarrow AB = \sqrt {16} = 4\left( {cm} \right)\\
+ A{C^2} = CH.BC = 1,8.5 = 9\\
\Rightarrow AC = \sqrt 9 = 3\left( {cm} \right)\\
2)\\
+ \sin \widehat B = \frac{{AC}}{{BC}} = \frac{3}{5} \Rightarrow \widehat B = {37^0}\\
\Rightarrow \widehat C = {90^0} - {37^0} = {53^0}\\
3)\\
Do:\widehat {ABD} = \widehat {CBD} = \frac{{\widehat B}}{2} = 18,{5^0}\\
\Rightarrow cos\widehat {ABD} = \frac{{AB}}{{BD}} = \frac{4}{{BD}}\\
\Rightarrow BD = \frac{4}{{cos18,{5^0}}} = 4,2\left( {cm} \right)\\
4)\\
Do\,BD\,là\,phân\,giác\,nên:\frac{{AD}}{{AB}} = \frac{{CD}}{{BC}} = \frac{{AD + CD}}{{AB + BC}} = \frac{{AC}}{{AB + BC}}\\
Mà:\tan \widehat {ABD} = \frac{{AD}}{{AB}}\\
\Rightarrow \tan \widehat {ABD} = \frac{{AC}}{{AB + BC}}
\end{array}$
(Áp dụng tính chất dãy tỉ số bằng nhau)
