If the sines of the angles of a triangle are in the ratio `3:5:7` their cotangents are in the ratio .....

Answer

To solve the problem of finding the ratio of the cotangents of the angles of a triangle when the sines of the angles are in the ratio \(3:5:7\), we can follow these steps:

### Step 1: Establish the relationship between the angles and their sines
Given that:
\[ \sin A : \sin B : \sin C = 3 : 5 : 7 \]
We can denote:
\[ \sin A = 3k, \quad \sin B = 5k, \quad \sin C = 7k \]
for some constant \(k\).

### Step 2: Use the sine rule
According to the sine rule:
\[ \frac{A}{\sin A} = \frac{B}{\sin B} = \frac{C}{\sin C} = 2R \]
where \(R\) is the circumradius of the triangle. This implies that the sides \(A\), \(B\), and \(C\) are proportional to the sines of the angles:
\[ A : B : C = \sin A : \sin B : \sin C = 3 : 5 : 7 \]
Thus, we can express the sides as:
\[ A = 3m, \quad B = 5m, \quad C = 7m \]
for some constant \(m\).

### Step 3: Calculate the cotangents
Now, we need to find the cotangents of the angles:
\[ \cot A = \frac{\cos A}{\sin A}, \quad \cot B = \frac{\cos B}{\sin B}, \quad \cot C = \frac{\cos C}{\sin C} \]

Using the cosine rule:
\[ \cos A = \frac{B^2 + C^2 - A^2}{2BC} \]
\[ \cos B = \frac{A^2 + C^2 - B^2}{2AC} \]
\[ \cos C = \frac{A^2 + B^2 - C^2}{2AB} \]

### Step 4: Substitute the sides into the cosine formulas
Substituting \(A = 3m\), \(B = 5m\), and \(C = 7m\):
1. For \(\cot A\):
\[ \cot A = \frac{\frac{(5m)^2 + (7m)^2 - (3m)^2}{2 \cdot 5m \cdot 7m}}{3k} = \frac{25m^2 + 49m^2 - 9m^2}{70m^2} \cdot \frac{1}{3k} \]
\[ = \frac{65m^2}{70m^2 \cdot 3k} = \frac{65}{210k} \]

2. For \(\cot B\):
\[ \cot B = \frac{\frac{(3m)^2 + (7m)^2 - (5m)^2}{2 \cdot 3m \cdot 7m}}{5k} = \frac{9m^2 + 49m^2 - 25m^2}{42m^2} \cdot \frac{1}{5k} \]
\[ = \frac{33m^2}{42m^2 \cdot 5k} = \frac{33}{210k} \]

3. For \(\cot C\):
\[ \cot C = \frac{\frac{(3m)^2 + (5m)^2 - (7m)^2}{2 \cdot 3m \cdot 5m}}{7k} = \frac{9m^2 + 25m^2 - 49m^2}{30m^2} \cdot \frac{1}{7k} \]
\[ = \frac{-15m^2}{30m^2 \cdot 7k} = \frac{-15}{210k} \]

### Step 5: Form the ratio of cotangents
Now we can form the ratio:
\[ \cot A : \cot B : \cot C = \frac{65}{210k} : \frac{33}{210k} : \frac{-15}{210k} \]
This simplifies to:
\[ 65 : 33 : -15 \]

### Final Answer
Thus, the ratio of the cotangents of the angles of the triangle is:
\[ \boxed{65 : 33 : -15} \]