Prove by the method of vectors that in a triangle `a/(sin A)=b/(sinB)=c/(sinC)`.

Answer

To prove that in a triangle \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \) using vectors, we can follow these steps:

### Step 1: Define the Triangle and Vectors
Consider a triangle with vertices \( A \), \( B \), and \( C \). Let the sides opposite to these vertices be denoted as \( a \), \( b \), and \( c \) respectively. We can represent the sides as vectors:
- Let \( \vec{A} \) be the vector from \( B \) to \( C \) (length \( a \)).
- Let \( \vec{B} \) be the vector from \( C \) to \( A \) (length \( b \)).
- Let \( \vec{C} \) be the vector from \( A \) to \( B \) (length \( c \)).

### Step 2: Use the Cross Product
We know that the sum of the vectors in a closed triangle is zero:
\[ \vec{A} + \vec{B} + \vec{C} = 0 \]
Taking the cross product of \( \vec{B} \) with both sides gives:
\[ \vec{A} \times \vec{B} + \vec{B} \times \vec{B} + \vec{C} \times \vec{B} = 0 \]
Since \( \vec{B} \times \vec{B} = 0 \), we simplify this to:
\[ \vec{A} \times \vec{B} + \vec{C} \times \vec{B} = 0 \]
This implies:
\[ \vec{A} \times \vec{B} = -\vec{C} \times \vec{B} \]

### Step 3: Magnitudes of the Cross Products
Taking the magnitudes of both sides, we have:
\[ |\vec{A} \times \vec{B}| = |\vec{C} \times \vec{B}| \]
Using the formula for the magnitude of the cross product, we can express this as:
\[ |\vec{A}| |\vec{B}| \sin(\theta_{AB}) = |\vec{C}| |\vec{B}| \sin(\theta_{CB}) \]
Where \( \theta_{AB} \) is the angle between vectors \( \vec{A} \) and \( \vec{B} \), and \( \theta_{CB} \) is the angle between vectors \( \vec{C} \) and \( \vec{B} \).

### Step 4: Substitute the Lengths and Angles
Substituting the lengths of the sides:
\[ a \cdot b \cdot \sin(C) = c \cdot b \cdot \sin(A) \]
Dividing both sides by \( b \):
\[ a \sin(C) = c \sin(A) \]
Thus, we can write:
\[ \frac{a}{\sin A} = \frac{c}{\sin C} \]

### Step 5: Repeat for Other Ratios
We can repeat this process for the other pairs:
1. Cross product \( \vec{C} \) with \( \vec{A} \) to show \( \frac{b}{\sin B} = \frac{c}{\sin C} \).
2. Cross product \( \vec{A} \) with \( \vec{C} \) to show \( \frac{a}{\sin A} = \frac{b}{\sin B} \).

### Conclusion
From these steps, we have shown that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]