It is an equilateral triangle so its angles are equal 60°. From the definition, we know that:
[tex]\sin60^\circ=\dfrac{h}{4}[/tex]
and
[tex]\sin60^\circ=\dfrac{\sqrt{3}}{2}[/tex]
so
[tex]\dfrac{\sqrt{3}}{2}=\dfrac{h}{4}\quad \Big|\cdot4\\\\\\h=\dfrac{4\cdot\sqrt{3}}{2}\\\\\boxed{h=2\sqrt{3}}\\[/tex]
