Đáp án:
b) $\widehat B \approx 53,{13^0};\widehat C \approx 36,{87^0}$
c) $AH = 24;BH = 18;CH = 32$
d) $ DB = \dfrac{{150}}{7};DC = \dfrac{{200}}{7}$
e) $BE = \dfrac{{45}}{2}$
Giải thích các bước giải:
a) Ta có:
${50^2} = {30^2} + {40^2} \Rightarrow B{C^2} = A{B^2} + A{C^2}$
$ \Rightarrow \Delta ABC$ vuông tại $A$.
b) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 30;AC = 40;BC = 50\\
\Rightarrow \left\{ \begin{array}{l}
\sin B = \dfrac{{AC}}{{BC}} = \dfrac{4}{5} \Rightarrow \widehat B \approx 53,{13^0}\\
\tan C = \dfrac{{AB}}{{AC}} = \dfrac{3}{4} \Rightarrow \widehat C \approx 36,{87^0}
\end{array} \right.
\end{array}$
Vậy $\widehat B \approx 53,{13^0};\widehat C \approx 36,{87^0}$
c) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 30;AC = 40;BC = 50\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}\\
\Rightarrow AH = \sqrt {\dfrac{1}{{\dfrac{1}{{{{30}^2}}} + \dfrac{1}{{{{40}^2}}}}}} = 24\\
BH = \dfrac{{A{B^2}}}{{BC}} = \dfrac{{{{30}^2}}}{{50}} = 18\\
CH = BC - BH = 50 - 18 = 32
\end{array} \right.
\end{array}$
Vậy $AH = 24;BH = 18;CH = 32$
d) Ta có:
$AD$ là phân giác của tam giác $ABC$ $ \Rightarrow \dfrac{{DB}}{{DC}} = \dfrac{{AB}}{{AC}} = \dfrac{3}{4}$
Mà $DB + DC = BC = 50$
$ \Rightarrow DB = \dfrac{{150}}{7};DC = \dfrac{{200}}{7}$
Vậy $ DB = \dfrac{{150}}{7};DC = \dfrac{{200}}{7}$
e) Ta có:
$\begin{array}{l}
\Delta ABH;\widehat {AHB} = {90^0};AB = 30;BH = 18;AH = 24\\
\Rightarrow \tan \widehat {BAH} = \dfrac{{BH}}{{AH}} = \dfrac{{18}}{{24}} = \dfrac{3}{4}\\
\Rightarrow \tan \widehat {BAE} = \dfrac{3}{4}\\
\Delta ABE;\widehat {ABE} = {90^0};AB = 30;\tan \widehat {BAE} = \dfrac{3}{4}\\
\Rightarrow \dfrac{{BE}}{{AB}} = \tan \widehat {BAE} = \dfrac{3}{4}\\
\Rightarrow BE = AB.\dfrac{3}{4} = 30.\dfrac{3}{4} = \dfrac{{45}}{2}
\end{array}$
Vậy $BE = \dfrac{{45}}{2}$
