Question

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

Explanation:

Step1: Recall angle - sum property of a triangle

The sum of the interior angles of a triangle is $180^{\circ}$, i.e., $m\angle A + m\angle B+m\angle C = 180^{\circ}$. Given $m\angle B = 130^{\circ}$.

Step2: Find the sum of the other two angles

Substitute $m\angle B = 130^{\circ}$ into the angle - sum formula: $m\angle A + 130^{\circ}+m\angle C = 180^{\circ}$. Then, $m\angle A + m\angle C=180^{\circ}- 130^{\circ}=50^{\circ}$.

Step3: Analyze the first option

For the first option $m\angle A + m\angle B$, since $m\angle B = 130^{\circ}$, if $m\angle A + m\angle B = 155^{\circ}$, then $m\angle A=155^{\circ}-130^{\circ}=25^{\circ}$, and $m\angle C = 25^{\circ}$ (because the triangle is isosceles). And $m\angle A + m\angle B=25^{\circ}+130^{\circ}=155^{\circ}$.

Answer:

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