We are dealing with a triangle $\triangle ABC$, where:

• $A$ and $B$ lie on the circle.
• $C$ is a point outside the circle.
• The line $AC$ is tangent to the circle at $A$.
• The line $BC$ is a secant intersecting the circle at point $X$.
• The arc $AB$ measures 176°, and angle $\angle CBA = 56^\circ$.

We are tasked with finding the measure of angle $\angle ACB$.

Step 1: Understand the relationship between the angle and the intercepted arc.

The key fact we use here is the inscribed angle theorem, which states that an angle formed by two chords in a circle is half of the measure of the intercepted arc. For instance:

• Angle $\angle ACB$ will be half of the measure of the arc subtended by it (arc $AB$).
• In addition, angle $\angle CBA$ is an angle outside the circle, which will have a relationship to the external angle theorem.

Step 2: Determine the measure of angle $\angle ACB$.

Given that the arc $AB = 176^\circ$, the angle subtended by the arc $AB$ at $C$, which is $\angle ACB$, is: $\angle ACB = \frac{1}{2} \times (360^\circ - 176^\circ) = \frac{1}{2} \times 184^\circ = 92^\circ.$ So, $\angle ACB = 92^\circ$.

Step 3: Use the external angle theorem.

The measure of $\angle CBA$ (56°) is an external angle to the triangle. According to the external angle theorem, the external angle is equal to the sum of the two non-adjacent interior angles: $\angle CBA = \angle ACB + \angle BAC.$ Given that $\angle CBA = 56^\circ$ and we have calculated $\angle ACB = 92^\circ$, we can use the equation to solve for $\angle BAC$: $56^\circ = 92^\circ + \angle BAC.$ Rearranging: $\angle BAC = 56^\circ - 92^\circ = -36^\circ.$ There seems to be a mistake, so please check my analysis again