Which of the following triangles demonstrates the Law of Sines by showing that the ratios of the lengths of sides to the sines of their opposite angles are equal ($\frac{a}{sin\left(A\right)}=\frac{b}{sin\left(B\right)}=\frac{c}{sin\left(C\right)}$)?

• A. A triangle where $\frac{a}{sin\left(A\right)}=\frac{b}{sin\left(B\right)}=\frac{c}{sin\left(C\right)}$
• B. A triangle where $a+b+c=180$
• C. A triangle where $a=b=c$
• D. A triangle where $A+B+C=90$

Answer: (A)

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. This can be written as: \(\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\).
2. Identify that this law relates the sides and angles of a triangle through these ratios, showing a proportional relationship rather than just sums or equalities of sides or angles.
3. Examine each option given: the first option explicitly shows the equality of these ratios, which matches the Law of Sines formula.
4. The other options describe different properties: the sum of sides equal to 180 is incorrect (it should be the sum of angles), equal sides imply an equilateral triangle but do not express the Law of Sines, and the sum of angles equal to 90 is incorrect for triangles (it should be 180 degrees).
5. Therefore, the triangle that demonstrates the Law of Sines is the one where \(\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\) holds true.