Answer
$BC \approx 14.0$ ft.
Work Step by Step
1. Find $\angle ABC = \angle ACB$
$= \frac{180-30}{2}$ This is an isosceles triangle that has the bottom two angles equal to each other
$= 75^{\circ}$
2. Use the Law of Sines to find $BC$
Let $x = BC$
$\frac{x}{sin(30)} = \frac{27}{sin(75)}$
$x = \frac{27sin(30)}{sin(75)}$
by GDC / calculator
$x = 13.976...$ ft
$x \approx 14.0$ ft
$BC \approx 14.0$ ft
