Answer: 1026 m (1.026 km)
Distance travelled along the slanting path:
\[200\times\frac{54}{3600}=3\ \text{km}=3000\ \text{m}\]
Let A be the starting point, C the touchdown point and B the point on the ground below A. The angle of depression \(20^\circ\) equals \(\angle ACB\) (alternate angles), and AC = 3000 m.
\[\sin20^\circ=\frac{AB}{AC}\Rightarrow AB=3000\times0.342=1026\ \text{m}\]