Chapter 2 - Acute Angles and Right Triangles - Section 2.5 Further Applications of Right Triangles - 2.5 Exercises - Page 90: 44b

$d = 48.3~ft$

We can find $\theta$: $\theta = \frac{57.3~S}{R}$ $\theta = \frac{(57.3)(485~ft)}{600~ft}$ $\theta = 46.3^{\circ}$ We can find the value of $d$: $d = R~(1-cos~\frac{\theta}{2})$ $d = (600~ft)(1-cos~\frac{46.3^{\circ}}{2})$ $d = 48.3~ft$