`ΔAHB_|_` tại `H`
`=>AB^2 = AH^2 + HB^2 = 4^2 + 20^2 = 416`
$\eqalign{ & \Rightarrow AB \approx 20,4 \cr & \tan \widehat {BAH} = {{HB} \over {HA}} = {{20} \over 4} = 5 \cr & \Rightarrow \,\,\,\,\widehat {BAH} \approx 78,{7^0} \cr & \Rightarrow \,\,\,\,\widehat {HAC} \approx 78,{7^0} + {45^0} \approx 123,{7^0} \cr}$
$\eqalign{ & \widehat {HAB} + \widehat {HBA} = {90^0} \cr & ⇒ \widehat {ABC} + \widehat {HBA} = {90^0} \cr & \Rightarrow \widehat {HAB} = \widehat {ABC} \cr & \Rightarrow \widehat {BCA} = {180^0} – \widehat {BAC} – \widehat {ABC} = {180^0} – \widehat {HAC} \cr}$
$\Rightarrow \,\,\,\,\widehat {BCA} \approx {180^0} – 123,{7^0} = 56,{3^0}.$
Ta có: ${{BC} \over {{\mathop{\rm s}\nolimits} {\rm{in4}}{5^0}}} = {{AB} \over {{\mathop{\rm s}\nolimits} {\rm{in56,}}{{\rm{3}}^0}}}$
$\Rightarrow \,\,BC = {{20,4} \over {{\mathop{\rm s}\nolimits} {\rm{in56,}}{{\rm{3}}^0}}}{\mathop{\rm s}\nolimits} {\rm{in4}}{5^0} \approx 17,4$
Vậy cây cao 17,4 m
