Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

Let $\triangle \mathrm{ABC}$ be the triangle such that $a=2, b=\sqrt{6}$ and $c=\sqrt{3}-1$.

Clearly, $b>a>c$. Then,

$\angle B$ is the greatest angle of $\triangle A B C$.  (Greatest side has greatest angle opposite to it)

Using cosine formula, we have

$\cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}$

$\Rightarrow \cos B=\frac{(\sqrt{3}-1)^{2}+2^{2}-(\sqrt{6})^{2}}{2 \times(\sqrt{3}-1) \times 2}$

$\Rightarrow \cos B=\frac{3+1-2 \sqrt{3}+4-6}{4(\sqrt{3}-1)}$

$\Rightarrow \cos B=\frac{2-2 \sqrt{3}}{4(\sqrt{3}-1)}=\frac{-2(\sqrt{3}-1)}{4(\sqrt{3}-1)}$

$\Rightarrow \cos B=-\frac{1}{2}=\cos 120^{\circ}$

$\Rightarrow B=120^{\circ}$

Hence, the measure of its greatest angle is 120º.