Let's break down the forces into their components and solve for \( T_1 \) and \( T_2 \). Given: - \( T_3 = 900 \) lbs - The downward force is 1000 lbs ## Step 1: Resolve Forces into Components ### Vertical (y) direction: \[ T_1 \sin(53^\circ) + T_2 \sin(45^\circ) - T_3 = 1000 \] ### Horizontal (x) direction: \[ T_1 \cos(53^\circ) - T_2 \cos(45^\circ) = 0 \] ## Step 2: Substitute Known Values \[ \sin(53^\circ) \approx 0.7986, \quad \cos(53^\circ) \approx 0.6018 \] \[ \sin(45^\circ) = 0.7071, \quad \cos(45^\circ) = 0.7071 \] ### y-direction: \[ T_1 (0.7986) + T_2 (0.7071) - 900 = 1000 \] \[ 0.7986 T_1 + 0.7071 T_2 = 1900 \] ### x-direction: \[ T_1 (0.6018) - T_2 (0.7071) = 0 \] \[ 0.6018 T_1 = 0.7071 T_2 \] \[ T_1 = \frac{0.7071}{0.6018} T_2 \] \[ T_1 \approx 1.174 T_2 \] ## Step 3: Solve the System Substitute \( T_1 \) into the y-equation: \[ 0.7986 (1.174 T_2) + 0.7071 T_2 = 1900 \] \[ 0.9372 T_2 + 0.7071 T_2 = 1900 \] \[ 1.6443 T_2 = 1900 \] \[ T_2 = \frac{1900}{1.6443} \approx 1155 \text{ lbs} \] Now, find \( T_1 \): \[ T_1 = 1.174 \times 1155 \approx 1357 \text{ lbs} \] --- ## **Final Answers** \[ \boxed{T_1 \approx 1357 \text{ lbs}} \] \[ \boxed{T_2 \approx 1155 \text{ lbs}} \]