Suppose a triangle has two sides of length 3 and 4, and the angle between these two sides is [tex][tex]$60^{\circ}$[/tex][/tex]. What is the length of the third side of the triangle?

A. [tex][tex]$\sqrt{13}$[/tex][/tex]
B. [tex][tex]$\sqrt{3}$[/tex][/tex]
C. [tex][tex]$4 \sqrt{3}$[/tex][/tex]
D. [tex][tex]$3$[/tex][/tex]



Answer :

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]