Đáp án:
a) $\widehat B \approx {53^0}7';\widehat C \approx {36^0}53'$
b) $AH = 9,6cm;BH = 7,2cm;CH = 12,8cm$
Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 12cm;AC = 16cm\\
\Rightarrow \tan B = \dfrac{{AC}}{{AB}} = \dfrac{4}{3}\\
\Rightarrow \widehat B \approx {53^0}7'\\
\Rightarrow \widehat C \approx {36^0}53'
\end{array}$
Vậy $\widehat B \approx {53^0}7';\widehat C \approx {36^0}53'$
b) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AH \bot BC = H;AB = 12cm;AC = 16cm\\
\Rightarrow \left\{ \begin{array}{l}
AH = \sqrt {\dfrac{1}{{\dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}}}} = 9,6cm\\
BC = \sqrt {A{B^2} + A{C^2}} = 20cm\\
BH = \dfrac{{A{B^2}}}{{BC}} = 7,2cm\\
CH = BC - BH = 12,8cm
\end{array} \right.
\end{array}$
Vậy $AH = 9,6cm;BH = 7,2cm;CH = 12,8cm$
c) Ta có:
$\begin{array}{l}
+ )\Delta AHB;\widehat H = {90^0};HM \bot AB = M\\
\Rightarrow A{H^2} = AM.AB\\
+ )\Delta AHC;\widehat H = {90^0};HN \bot AC = N\\
\Rightarrow A{H^2} = AN.AC
\end{array}$
Như vậy: $AM.AB=AN.AC$
d) Ta có:
$\begin{array}{l}
+ )\Delta AHB;\widehat H = {90^0};HM \bot AB = M\\
\Rightarrow HN = \dfrac{{AH.CH}}{{AC}}\\
+ )\Delta AHC;\widehat H = {90^0};HN \bot AC = N\\
\Rightarrow HM = \dfrac{{AH.BH}}{{AB}}\\
\Rightarrow HM.HN = \dfrac{{AH.CH}}{{AC}}.\dfrac{{AH.BH}}{{AB}}\\
\Rightarrow HM.HN = \dfrac{{A{H^2}.\left( {HB.HC} \right)}}{{AB.AC}}\\
\Rightarrow HM.HN = \dfrac{{A{H^2}.A{H^2}}}{{AH.BC}}\\
\Rightarrow HM.HN = \dfrac{{A{H^3}}}{{BC}}\\
\Rightarrow A{H^3} = HM.HN.BC
\end{array}$
