1. **Problem Statement:** Determine the average shearing stress in the pin at point C given two horizontal forces $P=12$ kip applied at pin B, and a pin diameter of $0.8$ in. 2. **Relevant Formula:** The average shearing stress $\tau$ in a pin subjected to shear force $F$ is given by: $$\tau = \frac{F}{A}$$ where $A$ is the cross-sectional area of the pin. 3. **Calculate the cross-sectional area $A$ of the pin:** The pin is circular with diameter $d=0.8$ in, so $$A = \frac{\pi d^2}{4} = \frac{\pi (0.8)^2}{4} = \frac{\pi \times 0.64}{4} = 0.5027 \text{ in}^2$$ 4. **Determine the shear force on the pin at C:** Since two horizontal forces $P=12$ kip act at pin B, and the pin at C connects the member BC, the shear force on pin C equals the force transmitted through member BC. From the geometry and force equilibrium (not fully detailed here), the shear force on pin C equals $P=12$ kip. 5. **Calculate the average shearing stress:** $$\tau = \frac{F}{A} = \frac{12}{0.5027} = 23.87 \text{ ksi}$$ **Final answer:** The average shearing stress in the pin at C is approximately **23.87 ksi**.