The measure of ∠ACB will be 110°
Explanation
According to the diagram below, [tex]DE[/tex] and [tex]DF[/tex] are the perpendicular bisectors of [tex]AC[/tex] and [tex]BC[/tex] respectively and they intersect side [tex]AB[/tex] at points [tex]P[/tex] and [tex]Q[/tex] respectively.
So, [tex]AE=CE[/tex] and [tex]CF=BF[/tex]
Now, according to the [tex]SAS[/tex] postulate, ΔAPE and ΔCPE are congruent each other. Also, ΔCFQ and ΔBFQ are congruent to each other.
That means, ∠PCE = ∠PAE and ∠FCQ = ∠FBQ
As ∠CPQ = 78° , so ∠PCE + ∠PAE = 78° or, ∠PCE = [tex]\frac{78}{2}=39[/tex] ° and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = [tex]\frac{62}{2}=31[/tex]°
Now, in triangle CPQ, ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°
Thus, ∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°
