Sides of a Δ are in ratio $1: \sqrt{3} : 2$, then angles of Δ are in ratio:

1 : 2 : 3

Let the sides a, b, c be the $1λ, \sqrt{3}λ$ and $2λ$ respectively.

Clearly $c^2 = a^2 + b^2$ $∴∠C=90°=\frac{π}{2};\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{λ^2(3+4-1)}{λ^24\sqrt{3}}=\frac{\sqrt{3}}{2}$

$∴ A = \frac{π}{6}$ and hence $B = \frac{π}{2}-A = \frac{π}{3}$ ∴ A : B : C = 1 : 2 : 3

Another Approach:

$\sin A : \sin B : \sin C =\frac{1}{2}:\frac{\sqrt{3}}{2}:1$  $⇒A=\frac{π}{6},B=\frac{π}{3}$ and $C=\frac{π}{2}$ ⇒ A : B : C = 1 : 2 : 3