Question

under which angle conditions could a triangle exist? check all that apply. 3 acute angles 2 acute angles, 1 right angle 1 acute angle, 1 right angle, 1 obtuse angle 1 acute angle, 2 obtuse angles 2 acute angles, 1 obtuse angle

Explanation:

To determine valid triangle angle conditions, we use the triangle angle - sum property (sum of interior angles = \(180^{\circ}\)) and definitions of angle types:

• Acute angle: Less than \(90^{\circ}\).
• Right angle: Equal to \(90^{\circ}\).
• Obtuse angle: Greater than \(90^{\circ}\) but less than \(180^{\circ}\).

1. 3 acute angles:

Let each acute angle be less than \(90^{\circ}\). For example, in an equilateral triangle, each angle is \(60^{\circ}\) (all acute), and \(60 + 60+60 = 180^{\circ}\), so a triangle with 3 acute angles is possible.

1. 2 acute angles, 1 right angle:

A right angle is \(90^{\circ}\). Let the two acute angles sum to \(90^{\circ}\) (since \(90 + \text{sum of two acutes}=180\), so sum of two acutes \( = 90^{\circ}\)). For example, in a right - angled triangle with angles \(90^{\circ}, 45^{\circ}, 45^{\circ}\), \(90 + 45+45=180^{\circ}\), so this is possible.

1. 1 acute angle, 1 right angle, 1 obtuse angle:

A right angle is \(90^{\circ}\) and an obtuse angle is greater than \(90^{\circ}\). The sum of a right angle and an obtuse angle will be greater than \(180^{\circ}\) (e.g., \(90 + 100+\text{acute}\gt180\) even if the acute angle is \(0^{\circ}\), which is not possible for a triangle). So this combination is not possible.

1. 1 acute angle, 2 obtuse angles:

Each obtuse angle is greater than \(90^{\circ}\), so the sum of two obtuse angles is greater than \(180^{\circ}\) (e.g., \(100 + 100+\text{acute}\gt180\) even if the acute angle is \(0^{\circ}\)). So this combination is not possible.

1. 2 acute angles, 1 obtuse angle:

An obtuse angle is greater than \(90^{\circ}\) but less than \(180^{\circ}\). Let the obtuse angle be \(100^{\circ}\), and the two acute angles sum to \(80^{\circ}\) (e.g., \(40^{\circ}\) and \(40^{\circ}\)). Then \(100 + 40+40 = 180^{\circ}\), so this is possible.

Answer:

A. 3 acute angles
B. 2 acute angles, 1 right angle
E. 2 acute angles, 1 obtuse angle