The force of the officer on the seat is the reaction force to the force of the seat on the officer. We analyze the forces on the officer. The seat exerts two forces on the officer: a vertical normal force $F_N$ and a horizontal centripetal force $F_c$. In the vertical direction, forces are balanced:

$$F_N = mg$$

In the horizontal direction, the force provides the centripetal acceleration $a_c = v^2/R$:

$$F_c = m a_c = m \frac{v^2}{R}$$

The net force from the seat on the officer, $F_{seat}$, is the vector sum of these components. Its magnitude is found using the Pythagorean theorem:

$$F_{seat} = \sqrt{F_N^2 + F_c^2} = \sqrt{(mg)^2 + \left(m \frac{v^2}{R}\right)^2} = m\sqrt{g^2 + \left(\frac{v^4}{R^2}\right)}$$

By Newton's third law, the magnitude of the force of the officer on the seat is the same. The angle $\theta$ this force makes with the vertical is given by:

$$\tan(\theta) = \frac{\text{horizontal component}}{\text{vertical component}} = \frac{F_c}{F_N} = \frac{m v^2/R}{mg} = \frac{v^2}{gR}$$ $$\theta = \arctan\left(\frac{v^2}{gR}\right)$$