Physics (10th Edition)

$0.528\; km$
Let $x$ = distance from base of arch $tan(20^{\circ}) = \frac{192}{x}$ $x = \frac{192}{tan(20^{\circ})}$ $x$ $\approx 527.52\; m$ $= 527.52\; m \times \frac{1\; km}{1000\; m}$ $= 0.528\; km$