According to the Law of Sines, which triangle below correctly demonstrates that the side opposite the larger angle is the larger side?

• A. A triangle with angles $60°$, $60°$, and $60°$, and all sides equal
• B. A triangle with angles $40°$, $60°$, and $80°$, and corresponding opposite sides $a$, $b$, $c$ where $c>b>a$
• C. A triangle with angles $40°$, $60°$, and $80°$, and corresponding opposite sides $a$, $b$, $c$ where $a>b>c$
• D. A triangle with angles $90°$, $45°$, and $45°$, and corresponding opposite sides $a$, $b$, $c$ where $a<b<c$

Answer: (B)

Verified step by step guidance

1. Recall the Law of Sines, which states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\).
2. Understand that this law implies the side opposite the larger angle will be longer because the sine of a larger angle (less than 180°) is greater, making the corresponding side length larger.
3. Examine the given triangles and their angles: for example, a triangle with angles 40°, 60°, and 80° has sides opposite these angles labeled \(a\), \(b\), and \(c\) respectively.
4. Apply the Law of Sines to compare side lengths: since \(\sin 80^\circ > \sin 60^\circ > \sin 40^\circ\), it follows that \(c > b > a\) for the sides opposite these angles.
5. Conclude that the triangle with angles 40°, 60°, and 80° and sides \(a\), \(b\), and \(c\) where \(c > b > a\) correctly demonstrates the Law of Sines relationship between larger angles and larger opposite sides.