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if isosceles triangle abc has a 130° angle at vertex b, which statement…

Question

if isosceles triangle abc has a 130° angle at vertex b, which statement must be true?
○ m∠a = 15° and m∠c = 35°
○ m∠a + m∠b = 155°
○ m∠a + m∠c = 60°
○ m∠a = 20° and m∠c = 30°

Explanation:

Step1: Recall triangle angle sum property

The sum of the interior angles of a triangle is \(180^{\circ}\). For \(\triangle ABC\), \(m\angle A + m\angle B + m\angle C=180^{\circ}\).

Step2: Analyze the isosceles triangle

In an isosceles triangle, the angles opposite equal sides are equal. Since \(\angle B = 130^{\circ}\), it cannot be one of the equal angles (because \(130^{\circ}+130^{\circ}=260^{\circ}>180^{\circ}\)), so \(\angle A=\angle C\).

Step3: Calculate \(m\angle A + m\angle C\)

We know \(m\angle B = 130^{\circ}\), and from the angle - sum property \(m\angle A + m\angle B + m\angle C = 180^{\circ}\). Substitute \(m\angle B\) into the formula: \(m\angle A + m\angle C+130^{\circ}=180^{\circ}\). Then \(m\angle A + m\angle C=180^{\circ}- 130^{\circ}=50^{\circ}\)? Wait, no, wait. Wait, if \(\angle A=\angle C\), then \(m\angle A=m\angle C=\frac{180 - 130}{2}=\frac{50}{2} = 25^{\circ}\). But let's check the options:

  • Option 1: \(m\angle A = 15^{\circ}\) and \(m\angle C = 35^{\circ}\). But in an isosceles triangle with \(\angle B = 130^{\circ}\), \(\angle A=\angle C\), so this is wrong.
  • Option 2: \(m\angle A + m\angle B\). We know \(m\angle A = 25^{\circ}\), \(m\angle B = 130^{\circ}\), so \(m\angle A + m\angle B=25 + 130=155^{\circ}\). Let's check other options.
  • Option 3: \(m\angle A + m\angle C\). Since \(m\angle A + m\angle C=180 - 130 = 50^{\circ}

eq60^{\circ}\), this is wrong.

  • Option 4: \(m\angle A = 20^{\circ}\) and \(m\angle C = 30^{\circ}\). But \(\angle A\) and \(\angle C\) should be equal, so this is wrong.

Wait, maybe I made a mistake in the first analysis. Let's re - do it. The sum of angles in a triangle is \(180^{\circ}\). Given \(\angle B = 130^{\circ}\), then \(\angle A+\angle C=180 - 130=50^{\circ}\)? No, wait, no. Wait, in an isosceles triangle, two angles are equal. If \(\angle B\) is the vertex angle (the angle between the two equal sides), then the other two angles (\(\angle A\) and \(\angle C\)) are equal. So \(\angle A=\angle C=\frac{180 - 130}{2}=25^{\circ}\). But let's check the options:

Option 2: \(m\angle A + m\angle B\). If \(\angle A = 25^{\circ}\), then \(m\angle A + m\angle B=25 + 130 = 155^{\circ}\). Let's check the other options:

Option 1: \(m\angle A = 15^{\circ}\), \(m\angle C = 35^{\circ}\). But \(\angle A\) and \(\angle C\) should be equal (since \(\angle B\) is \(130^{\circ}\), the other two angles are equal), so this is incorrect.

Option 3: \(m\angle A + m\angle C=180 - 130 = 50^{\circ}
eq60^{\circ}\), so this is incorrect.

Option 4: \(m\angle A = 20^{\circ}\), \(m\angle C = 30^{\circ}\). Again, \(\angle A\) and \(\angle C\) should be equal, so this is incorrect.

So the correct option is the second one: \(m\angle A + m\angle B = 155^{\circ}\)

Answer:

\(m\angle A + m\angle B = 155^{\circ}\) (the option with this statement)