Solution :
In triangle ABC, we have :
AB = \(6\sqrt{3}\) cm , AC = 12 cm
and BC = 6 cm
Now, \({AB}^2\) + \({BC}^2\) = \((6\sqrt{3})^2\) + \((6)^2\)
= \(36 \times 3\) + 36 = 108 + 36 = 144 = \((AC)^2\)
Thus, triangle ABC is a right angled triangle at B.
\(\therefore\) \(\angle\) B = 90
