Let's break down the problem step by step: ### **Given:** - \( |T_1| = 155 \) N, at an angle of \( 30^\circ \) above the \( x \)-axis. - \( |T_2| = 200 \) N, at an angle of \( 35^\circ \) below the \( x \)-axis. - We are to resolve the resultant force into \( x \) and \( y \) components. --- ### **Step 1: Resolve Each Force Into Components** #### **For \( T_1 \):** - \( T_1 \) is \( 30^\circ \) above the \( x \)-axis. \[ T_{1x} = |T_1| \cos(30^\circ) \] \[ T_{1y} = |T_1| \sin(30^\circ) \] Plug in the values: \[ T_{1x} = 155 \cos(30^\circ) = 155 \times 0.866 = 134.23 \text{ N} \] \[ T_{1y} = 155 \sin(30^\circ) = 155 \times 0.5 = 77.5 \text{ N} \] --- #### **For \( T_2 \):** - \( T_2 \) is \( 35^\circ \) below the \( x \)-axis (so the angle with respect to the positive x-axis is **-35°**). \[ T_{2x} = |T_2| \cos(-35^\circ) \] \[ T_{2y} = |T_2| \sin(-35^\circ) \] Plug in the values: \[ T_{2x} = 200 \cos(-35^\circ) = 200 \times 0.8192 = 163.84 \text{ N} \] \[ T_{2y} = 200 \sin(-35^\circ) = 200 \times (-0.5736) = -114.72 \text{ N} \] --- ### **Step 2: Add Components to Find Resultant Force** #### **Sum of x-components:** \[ R_x = T_{1x} + T_{2x} = 134.23 + 163.84 = 298.07 \text{ N} \] #### **Sum of y-components:** \[ R_y = T_{1y} + T_{2y} = 77.5 + (-114.72) = -37.22 \text{ N} \] --- ### **Final Answer:** \[ \boxed{ \begin{aligned} R_x &= 298.07~\text{N} \\ R_y &= -37.22~\text{N} \end{aligned} } \] --- #### **Explanation:** - Each force was split into its \( x \) and \( y \) components using trigonometry. - Angles above the \( x \)-axis are positive for \( y \), below are negative. - The components were summed to get the resultant \( x \) and \( y \) components.