Given triangle ABC with sides $a$, $b$, and $c$ opposite angles A, B, and C respectively, which equation can be used to solve for the measure of angle $B$?

• A. $b^2=a^2+c^2+2accos\left(B\right)$
• B. $b^2=a^2+c^2-2accos\left(B\right)$
• C. $c^2=a^2+b^2-2abcos\left(C\right)$
• D. $a^2=b^2+c^2-2bccos\left(A\right)$

Answer: (B)

Verified step by step guidance

1. Recognize that the problem involves the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles.
2. Recall the Law of Cosines formula for any triangle ABC: \(a^{2} = b^{2} + c^{2} - 2bc \cos(A)\), where side \(a\) is opposite angle \(A\), side \(b\) opposite angle \(B\), and side \(c\) opposite angle \(C\).
3. To solve for angle \(B\), use the Law of Cosines formula with side \(b\) opposite angle \(B\): \(b^{2} = a^{2} + c^{2} - 2ac \cos(B)\).
4. Note that the given incorrect equation has a plus sign before the \$2ac \cos(B)\( term, but the correct Law of Cosines formula requires a minus sign: \)b^{2} = a^{2} + c^{2} - 2ac \cos(B)$.
5. Therefore, the correct equation to solve for angle \(B\) is \(b^{2} = a^{2} + c^{2} - 2ac \cos(B)\), which can be rearranged to find \(\cos(B)\) and then \(B\) itself.