About the character degree of an extension I try to find some relations for the irreducible character degrees of the extensions of the groups. For example:
Let $ G $ be a finite group of order $1800$ such that $ G $ has a normal subgroup of order $30$ and $ G/N $ is isomorphic to the alternating group $ A_5$. How we  can prove that there is no irreducible character $\chi $ of $ G $ such that $15\mid \chi (1) $?
 A: The argument below is mainly group theoretic. I think it should be possible to give a more direct character-theoretic proof.
Let $H = O^{2}(N)$ which is cyclic of order $15$, and let $X = O^{2}(G).$ Then it is enough to prove that $X$ has no irreducible character of degree divisible by $15$, since $X \lhd G$ with $[G:X]$ a power of $2$, using Clifford's Theorem.
Note that $H \leq Z(X)$ since $H \lhd X$ and ${\rm Aut}(H)$ is a $2$-group.
Let $P$ be a Sylow $5$-subgroup of $X$. Then $P$ is Abelian and $P \cap X^{\prime} \cap Z(X) \leq P^{\prime} = 1$ by elementary transfer.  A similar argument works for the prime $3$. Hence $H \cap X^{\prime} =1$.
Now $XN/N \lhd G/N$ so either $X \leq N$ or $XN = G$. But if $X \leq N$ then $G$ is solvable, which is not the case. Hence $XN = G$ and $[G:X] = 2$.
Let $M$ be the terminal member of the derived series for $G$. Then $M \leq X$, so $M \leq X^{\prime}$. Hence $M \cap H = 1$ and $|M| \leq 60.$ Thus $M \cong A_{5}$ and $X \cong A_{5} \times H$, so all irreducible characters of $X$ have degree $1,3,4$ or $5$. 
A: Here is a more character-theoretic proof:  
As mentionend in the comments, $N$ has a normal cyclic subgroup $H$ of order $15$, which is characteristic in $N$ and thus normal in $G$. By Clifford's theorem, we have $\DeclareMathOperator{\Irr}{Irr}$
$$ \chi_H = e \sum_{g\in [G:T]} \lambda^g
  \quad\text{for some }
  \lambda\in \Irr(H),  $$
where $T=G_{\lambda}$ is the inertia group of $\lambda$. Since the Sylow $3$-subgroups and $5$-subgroups centralize $H$, the index $|G:T|$ divides $8$. Thus $15$ divides $e$, as $\chi(1) = e |G:T|$. By Frobenius reciprocity, $e = [\chi_H,\lambda] = [\chi, \lambda^G]$ and thus
$$ 8 \cdot 15 = |G:H| = \lambda^G(1) \geq e\chi(1) \geq 15^2,$$
contradiction. (The last argument just reproves the well-known fact that $e^2\leq |T:H|$.) 
Notice that I did not use the assumption $G/N = A_5$. 
