Relative angles

Answer: $60^\circ$

Finding angle B in terms of angle A and angle C

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

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

So, substituting in $B$ and $C$ in terms of $A$, $$\begin{array}{rl}A+\frac{3}{2}A+2A& =180\\ \Rightarrow \frac{9}{2}A& =180\\ \Rightarrow \frac{1}{2}A& =20\\ \Rightarrow A& =40\end{array}$$ So $B=\frac{3}{2}\times40=60$.

Finding angles A and C in terms of angle B

$ 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 $$\begin{array}{rl}\frac{2}{3}B+B+\frac{4}{3}B& =180\\ \Rightarrow 3B& =180\\ \Rightarrow B& =60\end{array}$$