Correct. One way to solve this triangle is to use the definition of sine to find x:

$\displaystyle\frac{{x}}{{6}}={{\sin{{33}}}}^{\circ}\approx{0.545}\Rightarrow{x}\approx{6}{\left({0.545}\right)}={3.27}\approx{3.3}\usepackage{cmbright}\Huge\$

To find y, you can either use another trigonometric function (such as cosine) or you can use the Pythagorean Theorem:

$\displaystyle{{y}}^{{2}}+{{3.3}}^{{2}}\approx{{6}}^{{2}}\Rightarrow{{y}}^{{2}}+{10.89}\approx{36}\Rightarrow{{y}}^{{2}}\approx{25.11}\Rightarrow{y}\approx{5.01}\approx{5.0}\usepackage{cmbright}\Huge\$

The two acute angles are complementary, so $\displaystyle{m}\angle{B}={{90}}^{\circ}-{{33}}^{\circ}={{57}}^{\circ}\usepackage{cmbright}\Huge\$ .