Question 378531
Suppose the vertices of the triangle are A,B,C such that AB = 22, angle C = 62 degrees (this will make labeling much easier). It follows that angle A = angle B = 59. By the law of sines,

{{{(AB)/(sin C) = (AC)/(sin B) = (BC)/(sin A)}}}

{{{22/(sin 62) = (AC)/(sin 59) = (BC)/(sin 59)}}}

Solving, we find AC = BC = {{{(22*sin(59))/(sin 62)}}} = 21.357...