A ball is held in equilibrium by three forces: T1, T2, and W. T1 acts upwards and to the right at an angle θ with the vertical. T2 acts horizontally to the left. W acts vertically downwards. Which equ

• $T_1$: Acts upwards and to the right, at an angle $\theta$ with the vertical.
• $T_2$: Acts horizontally to the left.
• $W$: Acts vertically downwards (weight of the ball).

Since the angle $\theta$ is given with respect to the vertical:

• Vertical component of $T_1$: $T_1 \cos \theta$ (acting upwards).
• Horizontal component of $T_1$: $T_1 \sin \theta$ (acting to the right).

For equilibrium, the sum of horizontal forces must be zero. Forces to the right must balance forces to the left.

For equilibrium, the sum of vertical forces must be zero. Forces upwards must balance forces downwards.

A. $T_1 \sin \theta = T_2$ (This matches our horizontal equilibrium equation).
B. $T_1 \cos \theta = T_2$ (Incorrect, $T_1 \cos \theta$ balances $W$, not $T_2$).
C. $T_1 + T_2 + W = 0$ (This represents the vector sum, but the options are scalar equations relating magnitudes).
D. $T_1 \sin \theta = W$ (Incorrect, $T_1 \sin \theta$ balances $T_2$, not $W$).

Therefore, the correct statement is $T_1 \sin \theta = T_2$.

The final answer is $\boxed{\text{A. } T_1 \sin \theta = T_2}$.