It looks like you've provided the problem statement, but the **image showing the directions and angles of members A and B is missing**. The directions (angles) are essential to solve the problem using trigonometry. However, I can **explain the general step-by-step method** for solving such problems, and you can plug in the correct angles from your diagram: --- ## Step-by-Step Solution ### 1. **Identify the Forces and Their Directions** - Let the force in member A: \( F_A = 10 \) kN - Let the force in member B: \( F_B = 15 \) kN - Assume angles \( \theta_A \) and \( \theta_B \) with respect to the horizontal (from the diagram). ### 2. **Resolve Each Force into Components** Assume: - \( F_A \) makes an angle \( \theta_A \) with the horizontal (counterclockwise from x-axis). - \( F_B \) makes an angle \( \theta_B \) with the horizontal. The components are: - \( F_{Ax} = F_A \cos(\theta_A) \) - \( F_{Ay} = F_A \sin(\theta_A) \) - \( F_{Bx} = F_B \cos(\theta_B) \) - \( F_{By} = F_B \sin(\theta_B) \) ### 3. **Add the Components** Sum the x-components and y-components: \[ R_x = F_{Ax} + F_{Bx} \] \[ R_y = F_{Ay} + F_{By} \] ### 4. **Find the Magnitude of the Resultant** \[ R = \sqrt{R_x^2 + R_y^2} \] ### 5. **Find the Direction of the Resultant** \[ \theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right) \] --- ## **Example Calculation (using hypothetical angles)** **Suppose:** - \( \theta_A = 30^\circ \) (above x-axis) - \( \theta_B = 120^\circ \) (from x-axis) ### 1. Components: - \( F_{Ax} = 10 \cos(30^\circ) = 10 \times .866 = 8.66 \) kN - \( F_{Ay} = 10 \sin(30^\circ) = 10 \times .5 = 5 \) kN - \( F_{Bx} = 15 \cos(120^\circ) = 15 \times (-.5) = -7.5 \) kN - \( F_{By} = 15 \sin(120^\circ) = 15 \times .866 = 12.99 \) kN ### 2. Add Components: - \( R_x = 8.66 + (-7.5) = 1.16 \) kN - \( R_y = 5 + 12.99 = 17.99 \) kN ### 3. Magnitude: \[ R = \sqrt{(1.16)^2 + (17.99)^2} \approx \sqrt{.13456 + 323.64} \approx \sqrt{323.77} \approx 17.99 \text{ kN} \] ### 4. Direction: \[ \theta_R = \tan^{-1}\left(\frac{17.99}{1.16}\right) \approx \tan^{-1}(15.51) \approx 86.3^\circ \] --- ## **Final Answer (with example angles):** - **Magnitude:** \( 17.99 \) kN - **Direction:** \( 86.3^\circ \) from the x-axis --- ### **Please provide the actual diagram or angles for your members**, and I can give you the precise answer for your problem!