A 30-60-90 triangle is half an equlateral triangle (refer to the image below).
If we call the length of the side [tex] \overline{BD} = x [/tex], we can see that it is exactly half of the side of the equilateral triangle. So, we have
[tex] \overline{AB} = \overline{AC} = \overline{BC} = 2\cdot \overline{BD} = 2x [/tex]
Moreover, we can find the height [tex] \overline{CD} [/tex] using the pythagorean theorem, having
[tex] \overline{CD} = \sqrt{4x^2-x^2} = \sqrt{3x^2} = x\sqrt{3} [/tex].
Now, you know the longer leg to be 6. The longer leg is the height, so you have
[tex] \overline{CD} = x\sqrt{3} = 6 \implies x = \overline{BD} = \cfrac{6}{\sqrt{3}} = \cfrac{6\sqrt{3}}{3} = 2\sqrt{3} [/tex]
So, the hypotenuse is twice that value:
[tex] BC = 2x = 2\cdot 2\sqrt{3} = 4\sqrt{3} [/tex]
