Lời giải:
Xét $\triangle ABC$ vuông tại C ,có : $\tan y=\frac{AC}{BC}=\frac{2}{5}=0,4$
=> $y\approx 21^{\circ}48{}'$
=> $x=90^{\circ}-y\approx 90^{\circ}- 21^{\circ}48{}'\approx 68^{\circ}12{}'$
=> $x-y\approx 68^{\circ}12{}'- 21^{\circ}48{}'\approx 46^{\circ}24{}'$
Vậy $x-y\approx 46^{\circ}24{}'$ .
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