Question

How would I solve this question

Answer 1

`tangent=(opposite)/(adjacent)\\\\\tan60^o=(y)/(8)\ \ \ |\tan60^o=\sqrt3\\\\(y)/(8)=\sqrt3\ \ \ \ |\cdot8\\\\\boxed{y=8\sqrt3}\to\boxed{y\approx13.9}`

Other method.

Two such triangles form an equilateral triangle (look at the picture).

The formula of a height of an equilateral triangle:

`h=(a\sqrt3)/(2)`

We have

`a=2\cdot8=16;\ h=y`

Substitute:

`y=(16\sqrt3)/(2)=8\sqrt3\approx13.9`

Answer 2

Tan(ANGLE) = Opposite Leg / Adjacent Leg

Tan(60) = Y/8

Y = 8 x tan(60)

Y = 8√3

Y = 13.9