We are given a triangle ABC with side lengths $AB = 7$ cm, $AC = 4$ cm, and the angle $BAC = 80^\circ$. We need to find the measure of angle $ACB$ to the nearest degree.

1. Problem Description

We are given a triangle ABC with side lengths $AB = 7$ cm, $AC = 4$ cm, and the angle $BAC = 80^\circ$. We need to find the measure of angle $ACB$ to the nearest degree.

2. Solution Steps

We can use the Law of Sines to find the angle $ACB$. Let $a = BC$, $b = AC = 4$ cm, and $c = AB = 7$ cm. The angle $A = BAC = 80^\circ$. We have

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

We are given $b$, $c$ and angle $A$. We want to find angle $C = ACB$. Using the Law of Sines,

$\frac{\sin C}{AB} = \frac{\sin A}{BC}$

We need to find $BC$ first. Using the Law of Cosines, we have

$a^2 = b^2 + c^2 - 2bc \cos A$

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

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

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

$a^2 = 65 - 9.7216$

$a^2 = 55.2784$

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

So $BC \approx 7.435$ cm. Now, we use the Law of Sines to find angle C.

$\frac{\sin C}{AB} = \frac{\sin A}{BC}$

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

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

$\sin C = \frac{7(0.9848)}{7.435}$

$\sin C = \frac{6.8936}{7.435}$

$\sin C \approx 0.9271$

$C = \arcsin(0.9271) \approx 67.9^\circ$

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

3. Final Answer

68