In solving triangles, the law of cosines is essential for determining unknown sides and angles, particularly in SAS (Side-Angle-Side) and SSS (Side-Side-Side) configurations. When given two sides and the included angle in an SAS triangle, the law of cosines allows us to find the third side. The formula used is:

$$c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$$

In this context, if we have a triangle where side a is 4, side b is 3, and angle C is 60 degrees, we can substitute these values into the formula to find side c:

$$c^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \cos(60^\circ)$$

Calculating this gives:

$$c^2 = 16 + 9 - 2 \cdot 4 \cdot 3 \cdot \frac{1}{2}$$

$$c^2 = 25 - 12 = 13$$

Thus, c is equal to the square root of 13, or approximately 3.6. However, it is often preferable to leave answers in radical form to avoid rounding errors.

Once the third side is found, the next step is to determine one of the remaining angles using the law of cosines again. For angle A, the formula is:

$$A^2 = b^2 + c^2 - 2bc \cdot \cos(A)$$

Substituting the known values:

$$4^2 = 3^2 + 13 - 2 \cdot 3 \cdot \sqrt{13} \cdot \cos(A)$$

Rearranging and simplifying leads to:

$$16 = 9 + 13 - 6\sqrt{13} \cdot \cos(A)$$

From here, we can isolate cos(A):

$$-6\sqrt{13} \cdot \cos(A) = -6$$

Thus, we find:

$$\cos(A) = \frac{1}{\sqrt{13}}$$

Taking the inverse cosine gives us angle A as approximately 73.9 degrees. Since the cosine function yields only one angle in the range of 0 to 180 degrees, no further calculations for a second angle are necessary.

Finally, to find the last angle B, we use the angle sum property of triangles:

$$A + B + C = 180^\circ$$

Substituting the known values:

$$73.9 + B + 60 = 180$$

Solving for B gives:

$$B = 180 - 73.9 - 60 = 46.1^\circ$$

In conclusion, the triangle's dimensions are complete with side c as the square root of 13, angle A as 73.9 degrees, and angle B as 46.1 degrees. Mastery of the law of cosines is crucial for effectively solving these types of triangle problems.