Step 1: Get the reactions from symmetry, same as before.
Because the geometry and the single 20 kN load at R both sit symmetric about the centreline through R, the pin at P and the roller at T each pick up 10 kN vertically, with no horizontal reaction at P.
Step 2: Apply Lami's theorem at joint P.
At joint P only three forces act: the vertical reaction $R_P = 10$ kN pointing up, the force in member PQ along the 60 degree incline, and the force in member PR along the horizontal. Treating these three concurrent forces with Lami's theorem, the angle between $R_P$ and PR is 90 degrees, the angle between $R_P$ and PQ is 150 degrees, and the angle between PQ and PR is 120 degrees.
Step 3: Write the Lami ratio and solve.
$\dfrac{R_P}{\sin(120^{\circ})} = \dfrac{F_{PR}}{\sin(150^{\circ})} = \dfrac{F_{PQ}}{\sin(90^{\circ})}$. Using $R_P = 10$ kN, $\sin(120^{\circ}) = \sqrt{3}/2$ and $\sin(150^{\circ}) = 1/2$:
$F_{PR} = R_P \times \dfrac{\sin(150^{\circ})}{\sin(120^{\circ})} = 10 \times \dfrac{1/2}{\sqrt{3}/2} = \dfrac{10}{\sqrt{3}}$ kN.
Final Answer:
This matches the joint resolution method exactly.
\[ \boxed{|F_{PR}| = \dfrac{10}{\sqrt{3}} \approx 5.77 \text{ kN}} \]
