How would I solve this question

Tan(ANGLE) = Opposite Leg / Adjacent Leg
Tan(60) = Y/8
Y = 8 x tan(60)
Y = 8√3
Y = 13.9
[tex]tangent=\dfrac{opposite}{adjacent}\\\\\tan60^o=\dfrac{y}{8}\ \ \ |\tan60^o=\sqrt3\\\\\dfrac{y}{8}=\sqrt3\ \ \ \ |\cdot8\\\\\boxed{y=8\sqrt3}\to\boxed{y\approx13.9}[/tex]
Other method.
Two such triangles form an equilateral triangle (look at the picture).
The formula of a height of an equilateral triangle:
[tex]h=\dfrac{a\sqrt3}{2}[/tex]
We have
[tex]a=2\cdot8=16;\ h=y[/tex]
Substitute:
[tex]y=\dfrac{16\sqrt3}{2}=8\sqrt3\approx13.9[/tex]