Let $\hat a$ be a unit vector and $\vec b$ a non-zero vector not parallel to $\hat a$. The angles of the triangle, two of whose sides are represented by $\sqrt{3} (\hat a×\vec b)$ and $\vec b-(\hat a.\vec b)\hat a$ are

$π/б, π/3, π/2$

We have,

$\hat a×(\vec b×\hat a)=(\hat a.\hat a)\vec b-(\hat a.\vec b)\hat a$

$⇒\hat a×(\vec b×\hat a)=\vec b-(\hat a.\vec b)\hat a$  $[∵\hat a.\hat a=|\hat a|^2=1]$

Thus, two sides of the triangle are represented by the vectors

$\sqrt{3}(\hat a×\vec b)$ and $\hat a×(\vec b×\hat a)$.

Since $\hat a×(\vec b×\hat a)$ is perpendicular to $\hat a$ and $\vec b×\hat a$. Therefore, angle between the sides represented by the given vectors is a right angle.

Now, 

$|\vec b-(\hat a×\vec b)\hat a|=\hat a×(\vec b×\hat a)$

$⇒|\vec b-(\hat a×\vec b)\hat a|=|\hat a||\vec b×\hat a|\sin π/2$  $[∵\hat a⊥(\vec b×\hat a)]$

$⇒|\vec b-(\hat a×\vec b)\hat a|=|\vec b×\hat a|=|\hat a×\vec b|$

Thus, the lengths of the sides containing the right angle of the triangle are in the ratio $1:\sqrt{3}$. So, the other two angles are $π/6$ and $π/3$.