Given, cyclic quadrilateral \[ABCD\]

So, \[\angle A\text{ }+\angle C\text{ }=\text{ }{{180}^{o}}~\][Opposite angles in a cyclic quadrilateral is supplementary]

\[3\angle C\text{ }+\angle C\text{ }=\text{ }{{180}^{o}}~\]

\[~[As\angle A\text{ }=\text{ }3\angle C]\]

\[\angle C\text{ }=\text{ }{{45}^{o}}\]

Now,

\[\angle A\text{ }=\text{ }3\angle C\text{ }=\text{ }3\text{ }x\text{ }{{45}^{o}}\]

\[\angle A\text{ }=\text{ }{{135}^{o}}\] 

Similarly,

\[\angle B\text{ }+\angle D\text{ }=\text{ }{{180}^{o}}\]  

\[[As\angle D\text{ }=\text{ }5\angle B]\]

\[\angle B\text{ }+\text{ }5\angle B\text{ }=\text{ }{{180}^{o}}\]

Or,

\[6\angle B\text{ }=\text{ }{{180}^{o}}\]

\[\angle B\text{ }=\text{ }{{30}^{o}}\]

Now,

\[\angle D\text{ }=\text{ }5\angle B\text{ }=\text{ }5\text{ }x\text{ }{{30}^{o}}\]

\[\angle D\text{ }=\text{ }{{150}^{o}}\]

Therefore,

\[\angle A\text{ }=\text{ }{{135}^{o}},\angle B\text{ }=\text{ }{{30}^{o}},\]

\[\angle C\text{ }=\text{ }{{45}^{o}},\angle D\text{ }=\text{ }{{150}^{o}}~\]