Question

under which condition could a triangle exist? 1 acute angle, 1 right angle, 1 obtuse angle 2 acute angles, 1 right angle 1 acute angle, 2 obtuse angles 1 right angle, 2 obtuse angles

Explanation:

Step1: Recall the sum of angles in a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). An acute angle is less than \(90^{\circ}\), a right angle is \(90^{\circ}\), and an obtuse angle is greater than \(90^{\circ}\).

Step2: Analyze each option

• Option 1: 1 acute angle, 1 right angle, 1 obtuse angle

Let the acute angle \(a<90^{\circ}\), right angle \(r = 90^{\circ}\), obtuse angle \(o>90^{\circ}\). Then \(a + r+o>90^{\circ}+90^{\circ}=180^{\circ}\).

• Option 2: 2 acute angles, 1 right angle

Let the acute angles be \(a_1\) and \(a_2\) (\(a_1<90^{\circ}\), \(a_2<90^{\circ}\)) and the right angle \(r = 90^{\circ}\). Then \(a_1 + a_2+r=a_1 + a_2+90^{\circ}\). Since \(a_1 + a_2<180^{\circ}\), it is possible (e.g., \(a_1 = 30^{\circ}\), \(a_2=60^{\circ}\), \(r = 90^{\circ}\), \(30^{\circ}+60^{\circ}+90^{\circ}=180^{\circ}\)).

• Option 3: 1 acute angle, 2 obtuse angles

Let the acute angle \(a<90^{\circ}\) and obtuse angles \(o_1>90^{\circ}\), \(o_2>90^{\circ}\). Then \(a+o_1 + o_2>90^{\circ}+90^{\circ}=180^{\circ}\).

• Option 4: 1 right angle, 2 obtuse angles

Let the right angle \(r = 90^{\circ}\) and obtuse angles \(o_1>90^{\circ}\), \(o_2>90^{\circ}\). Then \(r+o_1 + o_2>90^{\circ}+90^{\circ}=180^{\circ}\).

Answer:

2 acute angles, 1 right angle