Given triangle $ABC$, which equation can be used to find the measure of angle $BAC$ using the Law of Sines?

• A. $\frac{sin\left(BAC\right)}{a}=\frac{sin\left(ABC\right)}{b}=\frac{sin\left(ACB\right)}{c}$
• B. $\frac{cos\left(BAC\right)}{a}=\frac{cos\left(ABC\right)}{b}=\frac{cos\left(ACB\right)}{c}$
• C. $a=b^2+c^2-2bccos\left(BAC\right)$
• D. $\frac{a}{sin\left(BAC\right)}=\frac{b}{sin\left(ABC\right)}=\frac{c}{sin\left(ACB\right)}$

Answer: (A)

Verified step by step guidance

1. Recall the Law of Sines, which states that in any triangle ABC, the ratio of the length of a side to the sine of its opposite angle is constant. This can be written as \(\frac{a}{\sin(\angle BAC)} = \frac{b}{\sin(\angle ABC)} = \frac{c}{\sin(\angle ACB)}\).
2. Identify the angle you want to find, which is \(\angle BAC\), and the side opposite to it, which is side \(a\).
3. Set up the Law of Sines equation relating \(\sin(\angle BAC)\) and side \(a\) to another known angle and its opposite side, for example, \(\sin(\angle ABC)\) and side \(b\).
4. Write the equation as \(\frac{\sin(\angle BAC)}{a} = \frac{\sin(\angle ABC)}{b}\), which can be extended to include the third side and angle as \(\frac{\sin(\angle BAC)}{a} = \frac{\sin(\angle ABC)}{b} = \frac{\sin(\angle ACB)}{c}\).
5. Use this equation to solve for \(\sin(\angle BAC)\) if the sides and other angles are known, and then find the measure of \(\angle BAC\) by taking the inverse sine.