Convert the torque to SI units first and compute power in watts, then change to horsepower, as a cross-check on the direct formula. Torque \(T = 35\ \text{kgf-m} = 35 \times 9.81 = 343.35\ \text{N-m}\). Angular speed \(\omega = \dfrac{2\pi N}{60} = \dfrac{2\pi \times 1350}{60} = 141.37\ \text{rad/s}\).
Power \(P = T\omega = 343.35 \times 141.37 \approx 48540\ \text{W} = 48.54\ \text{kW}\).
Convert to horsepower using \(1\ \text{kW} = 1.341\ \text{HP}\):
\[\text{HP} = 48.54 \times 1.341 \approx 65.1\]
This matches the direct metric-horsepower calculation closely, confirming the result.
\[\boxed{\text{BHP} \approx 66}\]