Here is one way.

Draw the figure on paper so that you can follow.

In any Math exercise/problem, if there are givens, you normally find the answer by using those givens. It is just a matter of finding relationships among these givens, or how the givens relate to each other.

In your problem as posted, here are the givens:

angle PAC = angle AQC ------***

AO = OC ----------------------***

AQ = 108 cm. ------------***

CP = 75 cm. --------------***

And we are to find AC.

So looking for any relationships among those givens and the AC, we see the isosceles triangle AOC, and the two triangles PAC and AQC.

In triangle AOC:

AO = OC ----that is why triangle AOC is isosceles.

So, angle OAC = angle OCA ---------***

For less confusion, let

>>>angle PAC = angle AQC = angle theta

>>>angle OAC = angle OCA = angle U

>>>angle OCQ = angle V

>>>angle OPA = angle W

>>>side AC = x

Mark those down on your figure/drawing. -------****

In triangle PAC:

Sum of the interior angles of any triangle is 180 degrees,

U +theta +W = 180 ---(1)

By Law of Sines,

x/sinW = 75/sin(theta) ---(2)

In triangle AQC:

Three angles = 180 degrees,

U +theta +(U+V) = 180

2U +theta +V = 180 ------(3)

By Law of Sines,

x/sin(theta) = 108/sin(U+V) ----(4)

(3) minus (1),

2U +theta +V = 180 ------(3), minus...

U +theta +W = 180 ---(1), equals...

-----------------------

U +V -W = 0

So, (u+V) = W -----***

Substitute that into (4),

x/sin(theta) = 108/sinW -----(4a)

Now we see that (2) and (4a) have both sin(theta) and sinW.

From (2),

x/sinW = 75/sin(theta) ---(2)

Cross multiply,

x*sin(theta) = 75*sinW

sin(theta) = 75sinW / x ----***

From (4a),

x/sin(theta) = 108/sinW -----(4a)

Cross multiply,

x*sinW = 108*sin(theta)

sin(theta) = x*sinW / 108 ---***

sin(theta) = sin(theta),

75sinW / x = x*sinW /108

Cross multiply,

(75sinW)*108 = (x*sinW)*x

Divide both sides by sinW,

75*108 = x^2

x = sqrt(75*108) = sqrt(8100)

x = 90 cm. -------------------------answer.