Tam giác \(AHB\) vuông tại $H$ nên $AB^2=AH^2+HB^2=4^2+20^2=416$
\(\Rightarrow AB \approx 20,4\)
\(tan\widehat {BAH} = \frac{{HB}}{{HA}} = \frac{{20}}{4} = 5\\
\Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \widehat {BAH} \approx {78,7^0}\\
\Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \widehat {HAC} \approx {78,7^0} + {45^0} \approx {123,7^0}
\)
\( {\widehat {HAB} + \widehat {HBA} = {{90}^0}}\\ {\widehat {ABC} + \widehat {HBA} = {{90}^0}}\\ { \Rightarrow \widehat {HAB} = \widehat {ABC}}\\ { \Rightarrow \widehat {BCA} = {{180}^0}-\widehat {BAC}-\widehat {ABC} = {{180}^0}-\widehat {HAC}} \)
\( \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \widehat {BCA} \approx {180^0}-{123,7^0} = {56,3^0}.\)
Ta có: \(\frac{{BC}}{{{\rm{sin}}{{45}^0}}} = \frac{{AB}}{{{\rm{sin}}{{56,3}^0}}}\)
\(\Rightarrow {\mkern 1mu} {\mkern 1mu} BC = \frac{{20,4}}{{{\rm{sin}}{{56,3}^0}}}{\rm{sin}}{45^0} \approx 17,4\)
Vậy \(BC\approx17,4m\)
Tam giác \(AHB\) vuông tại H nên \(A{B^2} = A{H^2} + H{B^2} = {4^2} + {20^2} = 416\)
\(\Rightarrow AB \approx 20,4\)
\(tan\widehat {BAH} = \frac{{HB}}{{HA}} = \frac{{20}}{4} = 5\\
\Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \widehat {BAH} \approx {78,7^0}\\
\Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \widehat {HAC} \approx {78,7^0} + {45^0} \approx {123,7^0}
\)
\( {\widehat {HAB} + \widehat {HBA} = {{90}^0}}\\ {\widehat {ABC} + \widehat {HBA} = {{90}^0}}\\ { \Rightarrow \widehat {HAB} = \widehat {ABC}}\\ { \Rightarrow \widehat {BCA} = {{180}^0}-\widehat {BAC}-\widehat {ABC} = {{180}^0}-\widehat {HAC}} \)
\( \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \widehat {BCA} \approx {180^0}-{123,7^0} = {56,3^0}.\)
Ta có: \(\frac{{BC}}{{{\rm{sin}}{{45}^0}}} = \frac{{AB}}{{{\rm{sin}}{{56,3}^0}}}\)
\(\Rightarrow {\mkern 1mu} {\mkern 1mu} BC = \frac{{20,4}}{{{\rm{sin}}{{56,3}^0}}}{\rm{sin}}{45^0} \approx 17,4\)
Vậy \(BC\approx17,4m\)
