15) Given a triangle with an angle of 100° and an angle of α. Find the angle α.

Solution:

The image shows a triangle with a right angle (90°), an angle of 100°, and an angle labeled \( \alpha \). This diagram is geometrically impossible, as the sum of angles in a triangle must be 180°, and the presence of a 100° angle in a right-angled triangle is not possible.

Assuming the 100° is a typo and it should be an acute angle, or that it is not a right-angled triangle and the square symbol is incorrect, we cannot solve it without clarification.

If we ignore the right angle symbol and consider a triangle with angles 100° and \( \alpha \) and another unlabeled acute angle:

Let's assume the intention was a triangle where one angle is 100° and another is \( \alpha \), and the right angle symbol is incorrect. If the diagram were meant to represent a triangle with angles 100°, \( \alpha \) and, say, 40°, then \( \alpha = 180^{\circ} - 100^{\circ} - 40^{\circ} = 40^{\circ} \).

However, based STRICTLY on the image provided, it's an impossible triangle.

If we consider the 100° to be an exterior angle, or if the diagram has a mistake, we cannot proceed. Given the context of other problems, it is likely a typo. If we assume the angle indicated as 100° was meant to be acute, and the right angle is valid, then \( \alpha + 90^{\circ} + \text{acute angle} = 180^{\circ} \).

Given the provided solution format and the other problems, let's consider if the 100° angle is somehow related to \( \alpha \) in a different context or if it's a typo for an acute angle.

Let's assume there's a mistake in the problem and it's a right-angled triangle where one acute angle is \( \alpha \) and the other is related to 100° in some way (e.g., supplementary, making it 80°, which is still not a valid acute angle with \( \alpha \) to sum to 90°).

Let's consider the possibility that the 100° is a typo and it should be an angle that makes sense with a right triangle. For instance, if the other acute angle was 40°, then \( \alpha = 90^{\circ} - 40^{\circ} = 50^{\circ} \).

Without a valid geometric configuration, it is impossible to solve.

Answer: The problem as presented is geometrically impossible.