Check the picture below.
so let's use the law of sines then
[tex]\textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1.9}{sin(Z)}~~ = ~~\cfrac{XZ}{sin(Y)}\implies \cfrac{1.9}{sin(89^o)}~~ = ~~\cfrac{XZ}{sin(70^o)}\implies \cfrac{1.9\cdot sin(70^o)}{sin(89^o)}=XZ \\\\[-0.35em] ~\dotfill[/tex]
[tex]cos(21^o)=\cfrac{\stackrel{adjacent}{XW}}{\underset{hypotenuse}{XZ}}\implies XZ\cdot cos(21^o)=XW \\\\\\ \cfrac{1.9\cdot sin(70^o)\cdot cos(21^o)}{sin(89^o)}=XW\implies 1.7\approx XW[/tex]
quick observation, the picture of the triangle is very misleading, since those angles as drawn are not good representers of the angles values.
