



 
Hi Jesse, I drew a sketch of your pyramid. $R$ is the midpoint of the side $AB$ and $Q$ is at the centre of the square base. If this is a regular pyramid which I expect it is then the triangle $PQR$ is a right triangle. The length of $QR$ is 15" the length of $PQ$ is 18.25 inches and hence, from Pythagoras' Theorem \[RP^2 = PQ^2 + QR^2 = 18.25^2 + 15^2 = 558.0625 \mbox{ square inches }\] and hence $RP = \sqrt{558.0625} = 23.62$ inches. The side $ABP$ of the pyramid is an isosceles triangle and $R$ is the midpoint of $AB$ and hence triangle $RBP$ is a right triangle and Pythagoras' Theorem again gives is \[BP^2 = PR^2 + RB^2 = 23.62^2 + 15^2 = 783.06 \mbox{ square inches }\] an thus $BP = \sqrt{783.06} = 27.98$ inches. Again triangle $RBP$ is a right triangle, $RP = 15", PR = 23.62"$ and hence the measure of the angle $RBP$ is \[\cos^{1} \left(\frac{15}{23.62}\right) = 57.58^{o}.\] I hope this helps,  


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