$B=\frac{3}{4}C$ and $B =\frac{3}{2}A$, so $$\begin{align} \tfrac{3}{4}C&=\tfrac{3}{2}A \\ \Rightarrow \tfrac{1}{4}C&=\tfrac{1}{2}A \\ \Rightarrow C&=2A\end{align}$$.

Since $A$, $B$ and $C$ are the angles of a triangle, $A+B+C=180$.

So, substituting in $B$ and $C$ in terms of $A$, $$\begin{align}A+\tfrac32A+2A&=180\\

\Rightarrow \tfrac{9}{2}A&=180\\

\Rightarrow \tfrac{1}{2}A&=20\\

\Rightarrow A&=40\end{align}$$ So $B=\frac{3}{2}\times40=60$.

$ B=\frac{3}{4}C$ so $C=\frac{4}{3}B$.

And $B=\frac{3}{2}A$ so $A=\frac{2}{3}B$.

Since A, B and C are the angles of a triangle, $A+B+C=180$, so $$\begin{align}\tfrac{2}{3}B+B+\tfrac{4}{3}B&=180\\

\Rightarrow 3B&=180\\

\Rightarrow B&=60\end{align}$$

You can find more short problems, arranged by curriculum topic, in our short problems collection.