A triangle has sides in the ratio $1: \sqrt{2}: \sqrt{3}$. If the smallest angle is $\alpha^{\circ}$, what are the other two angles?

Let the sides be proportional to $1,\sqrt{2},\sqrt{3}$, so their squares are $1,2,3$. Since $1+2=3$, the triangle is right-angled by the converse of the Pythagorean theorem. The smallest side corresponds to the smallest angle $\alpha^{\circ}$, and the largest side corresponds to $90^{\circ}$. Therefore the third angle is $180^{\circ}-90^{\circ}-\alpha^{\circ}=(90-\alpha)^{\circ}$. Option D is tempting because side sizes increase, but angle measures do not become simple multiples here.