To determine the length of the third side of a triangle with sides of lengths 3 and 4 and an included angle of [tex]\(60^\circ\)[/tex], we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], and the angle [tex]\(\gamma\)[/tex] (where [tex]\(\gamma\)[/tex] is the angle opposite side [tex]\(c\)[/tex]):
[tex]\[ c^2 = a^2 + b^2 - 2ab \cos(\gamma) \][/tex]
Here, let [tex]\(a = 3\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(\gamma = 60^\circ\)[/tex].
### Step-by-Step Solution:
1. Convert the angle from degrees to radians:
[tex]\[
\gamma = 60^\circ = \frac{\pi}{3} \text{ radians} \approx 1.0471975511965976 \text{ radians}
\][/tex]
2. Calculate [tex]\(\cos(\gamma)\)[/tex]:
[tex]\[
\cos(60^\circ) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}
\][/tex]
3. Apply the Law of Cosines:
[tex]\[
c^2 = a^2 + b^2 - 2ab \cos(\gamma)
\][/tex]
Substituting the given values:
[tex]\[
c^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos\left(\frac{\pi}{3}\right)
\][/tex]
[tex]\[
c^2 = 9 + 16 - 2 \cdot 3 \cdot 4 \cdot \frac{1}{2}
\][/tex]
[tex]\[
c^2 = 25 - 12
\][/tex]
[tex]\[
c^2 = 13
\][/tex]
4. Solve for [tex]\(c\)[/tex] (the length of the third side):
[tex]\[
c = \sqrt{13} \approx 3.6055512754639887
\][/tex]
Based on these calculations, the length of the third side is [tex]\( \sqrt{13} \)[/tex]. Hence, the correct answer is:
A. [tex]\(\sqrt{13}\)[/tex]
