bearing in mind that a circle has a total of 360°, we can simply add those arcs and if, the arcs are equal in length, the legs of the inscribed triangle must also be equal, if on two are equals, will be the same for the triangle, and if none are equals, the same for the triangle.
[tex]\bf \stackrel{\widehat{QP}}{(8x-10)}+\stackrel{\widehat{PQ}}{(6x)}+\stackrel{\widehat{RQ}}{(10x+10)}~~=~~360\qquad \implies \qquad 24x=360 \\\\\\ x = \cfrac{360}{24}\implies x=15 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\widehat{QP}}{8(15)-10\implies 110}~\hfill \stackrel{\widehat{PQ}}{6(15)\implies 90}~\hfill \stackrel{\widehat{RQ}}{10(15)+10\implies 160}\\\\ ~\hfill \textit{\Large Scalene}~\hfill[/tex]
