Rectangles

Let W = Width and H = Height of the rectangle $\\$ From the Perimeter, we have $\\2(W+H)=26 \rightarrow W+H = 13 \rightarrow W=13-H \\$ From the Area, we have $\\W*H=42 \rightarrow (13-H)*H=42 \rightarrow 13H-H^2=42 \rightarrow H^2-13H+42=0 \rightarrow (H-6)(H-7)=0 \\$ In keeping with the diagram, let the Width (W) = 7 and the Height (H) = 6

From this, we calculate the length of the diagonal as $\\ |Diag| = \sqrt{W^2+H^2} = \sqrt{7^2+6^2} = \sqrt{85} \\$ and thus $\\Cos(\theta)=\frac{W}{|Diag|} = \frac{7}{\sqrt{85}}$ and $Sin(\theta) = \frac{H}{|Diag|} = \frac{6}{\sqrt{85}} \\$

We note that $P=2\theta$ so $\\Cos(P) = Cos(2\theta)=Cos^2(\theta)-Sin^2(\theta) = \frac{7^2}{85} - \frac{6^2}{85} = \frac{49-36}{85} = \boxed{\frac{13}{85}} \\$ (Note that if we had let the Width and Height be 6 and 7, the resulting value would simply have been negative.)