 Solution

Let $\varphi =(1+\sqrt{5})/2$ and $\bar{\varphi }=(1-\sqrt{5})/2$. In class order $(1,2A,3A,5A,5B)$, character orthogonality giveswith class sizes $1,15,20,12,12$. The rows are orthonormal and their squared degrees sum to $60$, so the table is complete.

A three-dimensional representation of $S_5$ would decompose into irreducibles. Since $S_5$ has irreducible degrees $1,1,4,4,5,5,6$, only one-dimensional constituents could occur, whose restrictions to $A_5$ are trivial. It cannot restrict to either irreducible $3$.