A different way to reach the same result is to work with the energy the glider trades for distance, instead of resolving forces along the flight path.
In a power-off glide, the only source of forward push is the loss of potential energy as the aircraft descends; that lost potential energy is used up fighting drag over the ground distance covered. For a small, steady glide angle $\gamma$, the drag force acting on the aircraft over a horizontal distance $R$ does work $D \times R$ (approximately, since the path stays close to horizontal for typical glide angles), and this must equal the potential energy given up, $W \times h$, where $h$ is the altitude lost:
\[ W h \approx D R \]Also, since the flight path is a straight glide, the lift very nearly supports the full weight, $L \approx W$ (true exactly when $\cos\gamma \approx 1$, and even in the exact treatment the ratio below is unaffected). Substituting $W \approx L$ into the energy balance:
\[ R \approx \frac{Lh}{D} = h\left(\frac{L}{D}\right) \]This shows the horizontal distance covered is the altitude lost multiplied by whatever lift to drag ratio the pilot flies at. To cover the maximum possible distance from a given height, the pilot must fly at the angle of attack that gives the maximum $L/D$, so:
\[ R_{max} = h \times \left(\frac{L}{D}\right)_{max} \]Putting in $h = 4$ km and $(L/D)_{max} = 15$:
\[ R_{max} = 4 \times 15 = 60 \text{ km} \]So the maximum glide range along the ground is 60 km, the same result as the force-balance method.