The sides of triangle are in A.P. and the greatest angle exceeds the least by $90°$ . These sides are in the ratio

Let $A=a-d,B=a,C=a+d$ be the angles of $\Delta ABC$

Given $C=90+A$

$\begin{array}{l}A+B+C=180\\ a-d+a+a+d=180\\ 3a=180\\ a=60\\ C=90+A\Rightarrow a+d=90+a-d\end{array}$

$\begin{array}{l}\Rightarrow 2d=90\\ \Rightarrow d=45\\ A=15;B=60;C=105\\ a:b:c=\sin A:\sin B:\sin C=\sin 15:\sin 60:\sin 105\end{array}$

$\begin{array}{l}=\frac{\sqrt{3}-1}{2\sqrt{2}}:\frac{\sqrt{3}}{2}:\frac{\sqrt{3}+1}{2\sqrt{2}}\\ =\sqrt{3}-1:\sqrt{6}:\sqrt{3}+1\end{array}$