Question

sophia wants to solve for the side length of ( b ). which equation could sophia use to solve for the side length of ( b ), given the other two side lengths and the measure of their included angle? ( b^2 = a^2 + c^2 - 2accos c ) ( b^2 = a^2 + c^2 - 2abcos c ) ( b^2 = a^2 + c^2 - 2accos a ) ( b^2 = a^2 + c^2 - 2accos b )

Explanation:

Step1: Recall 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. The formula is \( b^2 = a^2 + c^2 - 2ac\cos B \) when we want to find side \( b \) and the included angle is \( B \) (between sides \( a \) and \( c \)). Wait, let's check the triangle labeling. In the triangle, side \( b \) is opposite angle \( B \)? Wait, no, the triangle has vertices \( A \), \( B \), \( C \), with side \( a \) opposite \( A \), side \( b \) opposite \( B \), side \( c \) opposite \( C \)? Wait, no, looking at the diagram: side \( b \) is between \( C \) and \( A \), so the angle included between sides \( a \) and \( c \) is angle \( B \)? Wait, no, let's re-examine. The sides: side \( a \) is from \( C \) to \( B \), side \( c \) is from \( A \) to \( B \), and side \( b \) is from \( C \) to \( A \). So the included angle between sides \( a \) and \( c \) is angle \( B \) (at vertex \( B \)). Wait, no, the included angle for sides \( a \) and \( c \) (which are adjacent to angle \( B \)) would be angle \( B \). But the Law of Cosines formula for side \( b \) (opposite angle \( B \)) is \( b^2 = a^2 + c^2 - 2ac\cos B \). Wait, but let's check the options. Wait, the first option: \( b^2 = a^2 + c^2 - 2ac\cos C \). Let's see: angle \( C \) is at vertex \( C \), between sides \( a \) (from \( C \) to \( B \)) and \( b \) (from \( C \) to \( A \)). Wait, no, maybe I mixed up. Wait, the Law of Cosines is: for any triangle with sides \( a \), \( b \), \( c \) opposite angles \( A \), \( B \), \( C \) respectively, \( c^2 = a^2 + b^2 - 2ab\cos C \), \( a^2 = b^2 + c^2 - 2bc\cos A \), \( b^2 = a^2 + c^2 - 2ac\cos B \). Wait, but in the problem, we need to find side \( b \), given the other two sides \( a \) and \( c \), and their included angle. The included angle between sides \( a \) and \( c \) is angle \( B \) (since \( a \) is \( CB \), \( c \) is \( AB \), so the angle between them is at \( B \)). So the formula should be \( b^2 = a^2 + c^2 - 2ac\cos B \), which is the last option? Wait, no, let's check the options again. Wait, the options are:

1. \( b^2 = a^2 + c^2 - 2ac\cos C \)
2. \( b^2 = a^2 + c^2 - 2ab\cos C \)
3. \( b^2 = a^2 + c^2 - 2ac\cos A \)
4. \( b^2 = a^2 + c^2 - 2ac\cos B \)

Wait, maybe I mislabeled the sides. Let's look at the diagram: side \( b \) is \( CA \), side \( a \) is \( CB \), side \( c \) is \( AB \). So angle at \( C \) is between sides \( a \) (CB) and \( b \) (CA). Angle at \( A \) is between sides \( c \) (AB) and \( b \) (CA). Angle at \( B \) is between sides \( a \) (CB) and \( c \) (AB). So to find side \( b \) (CA), the two sides adjacent to the included angle: the sides forming the included angle with \( b \) would be \( a \) (CB) and \( c \) (AB)? No, wait, side \( b \) is between \( C \) and \( A \), so the included angle for sides \( a \) (from \( C \) to \( B \)) and \( c \) (from \( A \) to \( B \)) is angle \( B \) (at \( B \)). So the included angle between \( a \) and \( c \) is angle \( B \), so the Law of Cosines for side \( b \) (opposite angle \( B \)) is \( b^2 = a^2 + c^2 - 2ac\cos B \), which is the fourth option. Wait, but let's check the options again. Wait, the fourth option is \( b^2 = a^2 + c^2 - 2ac\cos B \). So that's the correct formula. Wait, but let's verify with the Law of Cosines. The Law of Cosines states that for a triangle with sides \( a \), \( b \), \( c \) opposite angles \( A \), \( B \), \( C \) respectively, \( b^2 = a^2 + c^2 - 2ac\cos B \). So in this case, side \( b \) is opposite angle \( B \), and the two sides forming angle \( B \) are \( a \) and \( c \) (since \( a \) is \( CB \), \( c \) is \( AB \), so angle \( B \) is between them). Therefore, the correct formula is \( b^2 = a^2 + c^2 - 2ac\cos B \), which is the fourth option.

Step2: Analyze the Options

- Option 1: \( b^2 = a^2 + c^2 - 2ac\cos C \). Angle \( C \) is at vertex \( C \), between sides \( a \) (CB) and \( b \) (CA). So this would be for side \( AB \) (side \( c \)), not \( b \). So incorrect.
- Option 2: \( b^2 = a^2 + c^2 - 2ab\cos C \). The term \( 2ab\cos C \) has \( ab \), but \( a \) and \( b \) are not the sides forming angle \( C \). Incorrect.
- Option 3: \( b^2 = a^2 + c^2 - 2ac\cos A \). Angle \( A \) is at vertex \( A \), between sides \( c \) (AB) and \( b \) (CA). So this would be for side \( BC \) (side \( a \)), not \( b \). Incorrect.
- Option 4: \( b^2 = a^2 + c^2 - 2ac\cos B \). Angle \( B \) is between sides \( a \) (CB) and \( c \) (AB), which are the sides adjacent to angle \( B \), and side \( b \) is opposite angle \( B \). So this matches the Law of Cosines formula. Correct.

Answer:

\( b^2 = a^2 + c^2 - 2ac\cos B \) (the fourth option: \( b^2 = a^2 + c^2 - 2ac\cos B \))