Question

10 multiple choice 5 points an exterior angle at the base of an isosceles triangle is always equal to the sum of the base angles greater than the sum of the base angles acute obtuse

Explanation:

Step1: Recall properties of isosceles triangle and exterior angles

In an isosceles triangle, the base angles are equal. Let the base - angle be \(x\). The exterior angle at the base is \(180 - x\).

Step2: Analyze each option

• Option 1: Sum of base angles is \(2x\). \(180 - x

eq2x\) (except when \(x = 60^{\circ}\), but not always).

• Option 2: \(180 - x>2x\) implies \(180>3x\) or \(x < 60^{\circ}\). But base angles of an isosceles triangle can be \(x\geq60^{\circ}\) (e.g., equilateral triangle \(x = 60^{\circ}\)).
• Option 3: If \(x\) is acute (\(x<90^{\circ}\)), then \(180 - x>90^{\circ}\) (obtuse). If \(x = 90^{\circ}\) (right - isosceles triangle), the exterior angle is \(90^{\circ}\).
• Option 4: Since the base angle \(x\) of an isosceles triangle is acute (\(x<90^{\circ}\) for non - right isosceles; in right - isosceles, the base angles \(x = 45^{\circ}\)), then the exterior angle \(180 - x\) is \(>90^{\circ}\) (obtuse).

Answer:

obtuse