Question

Given: KLIJ is inscribed in circle k(O)

Answer 1

Check the picture below.

let's notice that the angle at K is an inscribed angle with an intercepted arc

$\bf \stackrel{\textit{using the inscribed angle theorem}}{K=\cfrac{\widehat{LI}+\widehat{IJ}}{2}}\implies 9x+1=\cfrac{(10x-1)+59}{2} \\\\\\ 9x+1=\cfrac{10x+58}{2}\implies 18x+2=10x+58\implies 8x+2=58 \\\\\\ 8x=56\implies x=\cfrac{56}{8}\implies x=7 \\\\[-0.35em] ~\dotfill\\\\ K=9x+1\implies K=9(7)+1\implies \boxed{K=64}$

now, let's notice something again, the angle at L is also an inscribed angle, intercepting and arc of 97 + 59 = 156, so then, by the inscribed angle theorem,

∡L is half that, or 78°.

now, let's take a look at the picture down below, to the inscribed quadrilateral conjecture, since ∡J and ∡I are both supplementary angles, then

∡I = 180 - 64 = 116°.

∡J = 180 - 78 = 102°.