Like a lot of geometric figures, you can't really draw it until you find the solution. You can make an approximate drawing, then find the relationships necessary to solve the problem and create the scale drawing. The attached is a scale drawing.
Let's label some points, so we can talk about the relationships. Let D and E be the midpoints of AC and BC, respectively. Let the external point of intersection of PD and QE be point O, the center of the circle through A, B, and C.
Line CP is a reflection of line PQ across diameter line DO. That means ∠OPQ = ∠DPC = 90°-(78°/2) = 51°. Likewise, line CQ is a reflection of line PQ across diameter line EO. That means ∠OQP = ∠EQC = 90°-(62°/2) = 59°.
∠ACB is the supplement of ∠POQ, which is the supplement of the sum of ∠OPQ and ∠OQP. Thus ...
... ∠ACB = 51° +59° = 110°
_____
The perpendicular bisectors of the chords of a circle intersect at the circle center. For a triangle, the perpendicular bisectors of the sides intersect at the "circumcenter", the center of the circle in which the triangle is inscribed. Any point on the perpendicular bisector is the apex of an isosceles triangle whose base is the chord being bisected. Notions of symmetry and vertical angles get you to the same conclusions we described above regarding angles OPQ and DPC and their corresponding ones around point Q.
Of course, you remember that ∠ODC and ∠OEC are right angles, so angles ACB and POQ must be supplementary. (The angles of quadrilateral CDOE add to 360°.)