We are given a circle and three points $A, B, C$ on the circle which divide the circle into three arcs whose lengths are in the ratio $1:3:5$. We need to calculate the interior angles of triangle $ABC$.

1. Problem Description

We are given a circle and three points $A, B, C$ on the circle which divide the circle into three arcs whose lengths are in the ratio $1:3:5$. We need to calculate the interior angles of triangle $ABC$.

2. Solution Steps

Let the lengths of the arcs $AB$, $BC$, and $CA$ be $x, 3x, 5x$ respectively. The sum of the lengths of the arcs is the circumference of the circle, which corresponds to $360^\circ$.

Thus, $x + 3x + 5x = 360^\circ$.

$9x = 360^\circ$

$x = 40^\circ$

So the arc lengths are:

Arc $AB = x = 40^\circ$

Arc $BC = 3x = 3(40^\circ) = 120^\circ$

Arc $CA = 5x = 5(40^\circ) = 200^\circ$

Now, we use the property that the inscribed angle is half of the central angle subtended by the same arc. Also, the inscribed angle is half of the measure of the intercepted arc.

$\angle C = \frac{1}{2} \text{arc } AB = \frac{1}{2} (40^\circ) = 20^\circ$

$\angle A = \frac{1}{2} \text{arc } BC = \frac{1}{2} (120^\circ) = 60^\circ$

$\angle B = \frac{1}{2} \text{arc } CA = \frac{1}{2} (200^\circ) = 100^\circ$

3. Final Answer

The angles of triangle $ABC$ are $20^\circ, 60^\circ, 100^\circ$.