On circle there are points $A$, $B$ and $C$ such that they divide circle in ratio $3:5:7$. Find angles of triangle $ABC$
1. **Understanding the Problem:**
   We are given three points \( A \), \( B \), and \( C \) on a circle that divide the circle in the ratio \( 3:5:7 \). We need to find the angles of the triangle \( \triangle ABC \).

2. **Using the Central Angle Theorem:**
   If a segment \( XY \) of a circle centered at \( O \) cuts a portion \( x \) of its circumference, then the central angle \( \angle XOY \) is given by:
   \[
   \angle XOY = x \cdot 360^\circ
   \]
   For any point \( Z \) on the other side of the circle, the inscribed angle \( \angle XZY \) is half of the central angle:
   \[
   \angle XZY = \frac{1}{2} \angle XOY = x \cdot 180^\circ
   \]

3. **Applying the Ratios:**
   The points \( A \), \( B \), and \( C \) divide the circle in the ratio \( 3:5:7 \). This means the arcs subtended by these points are in the same ratio. Let the total circumference of the circle be \( 1 \) (for simplicity).

4. **Calculating the Angles:**
   - The arc corresponding to the ratio \( 3 \) is \( \frac{3}{15} \) of the total circumference.
   - The arc corresponding to the ratio \( 5 \) is \( \frac{5}{15} \) of the total circumference.
   - The arc corresponding to the ratio \( 7 \) is \( \frac{7}{15} \) of the total circumference.

   Using the inscribed angle theorem, the angles of the triangle \( \triangle ABC \) are:
   \[
   \angle A = \frac{3}{15} \cdot 180^\circ = 36^\circ
   \]
   \[
   \angle B = \frac{5}{15} \cdot 180^\circ = 60^\circ
   \]
   \[
   \angle C = \frac{7}{15} \cdot 180^\circ = 84^\circ
   \]

5. **Verification:**
   The sum of the angles in a triangle should be \( 180^\circ \):
   \[
   36^\circ + 60^\circ + 84^\circ = 180^\circ
   \]
   This confirms that our calculations are correct.

The final answer is \(\boxed{36^\circ, 60^\circ, 84^\circ}\).