In triangle $\mathrm{ABC}$, if side $\mathrm{AB}$ has length $b$, side $\mathrm{BC}$ has length $a$, and the angle at vertex $C$ is $\gamma$, what is the length of side $\mathrm{AC}$ according to the Law of Cosines?

• A. $\sqrt{a^2+b^2+2abcos\left(\gamma \right)}$
• B. $\sqrt{a^2+b^2-2absin\left(\gamma \right)}$
• C. $\sqrt{a^2+b^2-2abcos\left(\gamma \right)}$
• D. $\sqrt{a^2+b^2}$

Answer: (C)

Verified step by step guidance

1. Identify the sides and angle given in the triangle: side AB has length \( b \), side BC has length \( a \), and the angle at vertex C is \( \gamma \). We want to find the length of side AC.
2. Recall the Law of Cosines formula, which relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides \( a, b, c \) opposite angles \( A, B, C \) respectively, the Law of Cosines states: \[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
3. In this problem, side AC is opposite angle \( \gamma \) at vertex C, so we assign: \[ AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\gamma) \]
4. Substitute the given side lengths into the formula: \[ AC^2 = b^2 + a^2 - 2ab \cos(\gamma) \]
5. To find the length of side AC, take the square root of both sides: \[ AC = \sqrt{a^2 + b^2 - 2ab \cos(\gamma)} \]