Mark the correct alternative in each of the following:

Let $\triangle \mathrm{ABC}$ be the given triangle such that its sides are in the ratio $1: \sqrt{3}: 2$.

$\therefore a=k, b=\sqrt{3} k, c=2 k$

Now, $a^{2}+b^{2}=k^{2}+3 k^{2}=4 k^{2}=c^{2}$

So, ∆ABC is a right triangle right angled at C.

$\therefore C=90^{\circ}$

Using sine rule, we have

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\Rightarrow \frac{k}{\sin A}=\frac{\sqrt{3} k}{\sin B}=\frac{2 k}{\sin 90^{\circ}}$

$\Rightarrow \sin A=\frac{1}{2}$ and $\sin B=\frac{\sqrt{3}}{2}$

$\Rightarrow A=30^{\circ}$ and $B=60^{\circ}$

Thus, the measure of its greatest angle is $\frac{\pi}{2}$.

Hence, the correct answer is option (c)