The work-energy principle states that the work done on the train is equal to the change in kinetic energy plus the change in potential energy.

The change in kinetic energy (KE) is given by:

\(\Delta KE = \frac{1}{2} \times 150,000 \times 42^2 - \frac{1}{2} \times 150,000 \times 45^2 = -19,575,000 \text{ J}\)

The potential energy (PE) gain is:

\(\Delta PE = 150,000g \times 500 \sin \alpha\)

The total work done by the driving force and against resistance is:

\(16,000 \times 500 - 4 \times 10^6\)

Using the work-energy equation:

\(150,000g \times 500 \sin \alpha = 19,575,000 + 16,000 \times 500 - 4 \times 10^6\)

Solving for \(\sin \alpha\):

\(150,000g \times 500 \sin \alpha = 19,575,000 + 8,000,000 - 4,000,000\)

\(150,000g \times 500 \sin \alpha = 23,575,000\)

\(\sin \alpha = \frac{23,575,000}{150,000 \times 500 \times 9.8}\)

\(\alpha = \arcsin(0.0314) \approx 1.8^\circ\)