The inscribed angle theorem says angle ACB has measure equal to half the intercepted arc, so
m∠ACB = 1/2 × 116° = 58°
Triangle ABC is isosceles (since AC and BC are congruent, per the tick mark). This means angles CAB and ABC are both congruent. The sum of the interior angles to any triangle is 180°, so
m∠ACB + m∠ABC + m∠CAB = 180°
58° + 2 m∠ABC = 180°
2 m∠ABC = 122°
m∠ABC = 61°
Applying the inscribed angle theroem again tells you that arcs AC and BC have twice the measure of angle ABC/CAB, so they are both 122°.
The major arc ACB is the sum of arcs AC and BC, so its measure is 2 × 122° = 244°.
To recap:
arc AC = 122°
arc ACB = 244°
m∠ACB = 58°