Let's solve the problem step by step: Given: - \( P = 365 \) lb (force P) - \( Q = 220 \) lb (force Q) - Q makes a \( 40^\circ \) angle with the positive x-axis. - P is in the second quadrant, direction given by a 3-4-5 triangle (i.e., angle with the negative x-axis). Let's break each force into components. --- ### **Step 1: Resolve Force Q** Force Q is \( 40^\circ \) above the positive x-axis. - \( Q_x = Q \cos(40^\circ) \) - \( Q_y = Q \sin(40^\circ) \) Plug in values: \[ Q_x = 220 \cos(40^\circ) = 220 \times 0.7660 = 168.52 \text{ lb} \] \[ Q_y = 220 \sin(40^\circ) = 220 \times 0.6428 = 141.42 \text{ lb} \] --- ### **Step 2: Resolve Force P** From the triangle on the diagram: - The direction of P is up and to the left (second quadrant). - The components are: - \( x \): negative (to the left) - \( y \): positive (up) - From the triangle, for every 3 units left (x), 4 units up (y), the hypotenuse is 5 units. - So, - \( \cos(\theta_P) = \frac{-3}{5} \) - \( \sin(\theta_P) = \frac{4}{5} \) \[ P_x = P \cos(\theta_P) = 365 \times \left(-\frac{3}{5}\right) = 365 \times -0.6 = -219 \text{ lb} \] \[ P_y = P \sin(\theta_P) = 365 \times \frac{4}{5} = 365 \times 0.8 = 292 \text{ lb} \] --- ### **Step 3: Find the Resultant Components** Sum the x and y components: \[ R_x = P_x + Q_x = (-219) + (168.52) = -50.48 \text{ lb} \] \[ R_y = P_y + Q_y = 292 + 141.42 = 433.42 \text{ lb} \] --- ### **Step 4: Find the Magnitude of the Resultant \( R \)** \[ R = \sqrt{R_x^2 + R_y^2} \] \[ R = \sqrt{(-50.48)^2 + (433.42)^2} \] \[ R = \sqrt{2548.19 + 187857.51} \] \[ R = \sqrt{190405.7} \] \[ R \approx 436.4 \text{ lb} \] --- ### **Step 5: Find the Angle \( \theta \) with the Positive x-axis** \[ \tan \theta = \frac{R_y}{R_x} \] \[ \theta = \tan^{-1} \left(\frac{433.42}{-50.48}\right) \] \[ \theta = \tan^{-1}(-8.59) \] This gives an angle in the second quadrant (since \( R_x \) is negative, \( R_y \) is positive): \[ \theta \approx 96.6^\circ \] --- ## **Final Answers** ### (a) The resultant \( R \) of the two forces is: \[ \boxed{436.4 \text{ lb}} \] ### (b) The angle \( \theta \) the resultant makes with the positive x-axis is: \[ \boxed{96.6^\circ} \] (measured counterclockwise from the positive x-axis) Let me know if you want the answers in a different format or need a diagram!