Volume

Solve for $x$ by cosine law,

$6^2 = x^2 + x^2 - 2(x)(x) cos$ $120$

$x = 2\sqrt{3}$

Solve for $h$ by Pythagorean Theorem

$h = \sqrt{8^2 - x^2} = \sqrt{64 - 12} = \sqrt{52}$

Solve for the area of the base,

$A_{b} = \frac{1}{2}(6^2)(sin$ $60) = 18$ $sin$ $60$

Now solve for the volume,

$V = \frac{1}{3} A_{b}h = \frac{1}{3}(18$ $sin$ $60)(\sqrt{52}) = 37$ $cubic$ $units$