Suppose a triangle has two sides of length 3 and 4 and that the angle between these...

1. Identify the problem type: This problem involves finding the length of a side of a triangle given two sides and the included angle. This is a trigonometry problem that can be solved using the Law of Cosines.

1. State the Law of Cosines: The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides $a$, $b$, and $c$, and the angle $C$ opposite side $c$, the law is given by: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$

1. Assign variables: Let the two given sides be $a = 3$ and $b = 4$. Let the angle between them be $C = 60^\circ$. We need to find the length of the third side, $c$.

4. Substitute values into the Law of Cosines:
$$c^2 = 3^2 + 4^2 - 2(3)(4) \cos(60^\circ)$$

5. Calculate the squares of the sides:
$$c^2 = 9 + 16 - 2(3)(4) \cos(60^\circ)$$

6. Calculate the product of the sides and the factor of 2:
$$c^2 = 9 + 16 - 24 \cos(60^\circ)$$

7. Find the cosine of the angle: The cosine of $60^\circ$ is $\frac{1}{2}$.
$$c^2 = 9 + 16 - 24 \left(\frac{1}{2}\right)$$

8. Perform the multiplication:
$$c^2 = 9 + 16 - 12$$

9. Perform the addition and subtraction:
$$c^2 = 25 - 12$$
$$c^2 = 13$$

10. Solve for c by taking the square root: Since $c$ represents a length, it must be positive.
$$c = \sqrt{13}$$

1. State the final answer: The length of the third side of the triangle is $\sqrt{13}$.