The suspended object has two forces acting on it: gravity ($mg$) acting vertically downward and tension ($T$) acting along the rope at an angle $\theta$ to the vertical. The object shares the same horizontal acceleration $a$ as the train. We apply Newton's second law by resolving forces into horizontal (x) and vertical (y) components.

In the vertical direction, there is no acceleration:

$$\sum F_y = T \cos\theta - mg = 0$$ $$T = \frac{mg}{\cos\theta}$$

In the horizontal direction, the net force causes the acceleration $a$:

$$\sum F_x = T \sin\theta = ma$$

To find the acceleration $a$, substitute the expression for $T$ from the vertical component equation into the horizontal component equation:

$$a = \frac{T \sin\theta}{m} = \frac{1}{m} \left( \frac{mg}{\cos\theta} \right) \sin\theta$$ $$a = g \frac{\sin\theta}{\cos\theta} = g \tan\theta$$