A pilot flying over a level stretch of Saskatchewan sees two towns, which he knows to be 12 km apart. The angle of depression to the Town A is 56 degrees and the angle of depression to Town B is 32 degrees. Find the distance between the plane and Town A, round to the nearest tenth of a km.
Let $x$ be the distance between the plane and Town A. We then have $\tan 56^\circ = \frac{x}{12}$, so $x = \tan 56^\circ \cdot 12$ (since opposite over adjacent). Using our calculator, $\tan 56^\circ \approx 1.540,\!739$, so $x \approx 1.540,\!739 \cdot 12 = \boxed{18.5}$ km.
