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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 the properties of an isosceles triangle and the sum of angles in a triangle

In an isosceles triangle, two angles are equal. The sum of the interior angles of a triangle is \(180^{\circ}\). Given \(\angle B = 130^{\circ}\). Since \(130^{\circ}\) is an obtuse angle, it cannot be one of the equal angles (because if there were two angles of \(130^{\circ}\), \(130 + 130=260>180\)). So \(\angle A=\angle C\).

Step2: Calculate \(\angle A\) and \(\angle C\)

Let \(\angle A = \angle C=x\). Using the angle - sum formula for a triangle: \(\angle A+\angle B+\angle C = 180^{\circ}\). Substitute \(\angle B = 130^{\circ}\) and \(\angle A=\angle C=x\) into the formula: \(x + 130+x=180\). Combine like terms: \(2x=180 - 130\). So \(2x = 50\), and \(x = 25^{\circ}\).

Step3: Check each option

  • Option 1: \(m\angle A = 15^{\circ}\) and \(m\angle C = 35^{\circ}\). Since \(\angle A=\angle C\), this is false.
  • Option 2: \(m\angle A+m\angle B\). Since \(m\angle A = 25^{\circ}\) and \(m\angle B=130^{\circ}\), \(m\angle A + m\angle B=25 + 130=155^{\circ}\).
  • Option 3: \(m\angle A+m\angle C\). Since \(m\angle A=m\angle C = 25^{\circ}\), \(m\angle A + m\angle C=25+25 = 50^{\circ}

eq60^{\circ}\).

  • Option 4: \(m\angle A = 20^{\circ}\) and \(m\angle C = 30^{\circ}\). Since \(\angle A=\angle C\), this is false.

Answer:

\(m\angle A + m\angle B=155^{\circ}\)