The problem asks to construct a triangle $ABC$ with $AB = 7$ cm, $AC = 4$ cm, and angle $BAC = 80^\circ$. After constructing the triangle, we are asked to measure the size of angle $ACB$ to the nearest degree. Since we do not have the tools to construct and measure accurately, we will use the Law of Sines to solve the problem.

1. Problem Description

The problem asks to construct a triangle $ABC$ with $AB = 7$ cm, $AC = 4$ cm, and angle $BAC = 80^\circ$. After constructing the triangle, we are asked to measure the size of angle $ACB$ to the nearest degree. Since we do not have the tools to construct and measure accurately, we will use the Law of Sines to solve the problem.

2. Solution Steps

We can use the Law of Cosines to find the length of side $BC$. Let $a$ be the length of $BC$, $b$ be the length of $AC$, and $c$ be the length of $AB$. Then we have $a^2 = b^2 + c^2 - 2bc \cos(A)$, where $A$ is the angle $BAC$.

$a^2 = 4^2 + 7^2 - 2(4)(7) \cos(80^\circ)$

$a^2 = 16 + 49 - 56 \cos(80^\circ)$

$a^2 = 65 - 56 \cos(80^\circ)$

Using a calculator, $\cos(80^\circ) \approx 0.1736$.

$a^2 = 65 - 56(0.1736) = 65 - 9.7216 = 55.2784$

$a = \sqrt{55.2784} \approx 7.435$ cm.

Now, we use the Law of Sines to find the angle $ACB$, which we will denote as $C$.

$\frac{a}{\sin(A)} = \frac{c}{\sin(C)}$

$\frac{7.435}{\sin(80^\circ)} = \frac{7}{\sin(C)}$

$\sin(C) = \frac{7 \sin(80^\circ)}{7.435} = \frac{7(0.9848)}{7.435} = \frac{6.8936}{7.435} \approx 0.9272$

$C = \arcsin(0.9272) \approx 67.97^\circ$

To the nearest degree, $C \approx 68^\circ$.

The sum of angles in a triangle is $180^\circ$. Therefore, angle $ABC = B$ is given by:

$B = 180^\circ - A - C = 180^\circ - 80^\circ - 68^\circ = 32^\circ$.

3. Final Answer

The size of angle $ACB$ is approximately 68 degrees.