Answer :
To determine the true statement when given an isosceles triangle \(ABC\) with a \(130^\circ\) angle at vertex \(B\):
1. Identify the Given Information:
- \(\angle B = 130^\circ\)
- \(\triangle ABC\) is isosceles, implying that the two base angles \(\angle A\) and \(\angle C\) are equal.
2. Recall the Property of Triangle Angles Sum:
- The sum of the angles in any triangle is always \(180^\circ\).
3. Calculate the Base Angles:
- Since \(\triangle ABC\) is isosceles, the base angles \(\angle A\) and \(\angle C\) are equal.
- Let’s denote these angles as \(\angle A = \angle C\).
- The total sum of angles in the triangle is \(180^\circ\), so:
[tex]\[ \angle A + \angle B + \angle C = 180^\circ \][/tex]
- Substitute \(\angle B = 130^\circ\):
[tex]\[ \angle A + 130^\circ + \angle C = 180^\circ \][/tex]
Since \(\angle A = \angle C\), we can write:
[tex]\[ 2\angle A + 130^\circ = 180^\circ \][/tex]
4. Solve for \(\angle A\):
- Subtract \(130^\circ\) from both sides:
[tex]\[ 2\angle A = 50^\circ \][/tex]
- Divide both sides by 2:
[tex]\[ \angle A = \angle C = 25^\circ \][/tex]
5. Check the Provided Statements:
- \( m_{\angle A}=15^\circ \) and \( m_{\angle C}=35^\circ \) is incorrect, as both \(\angle A\) and \(\angle C\) are \(25^\circ\).
- \( m_{\angle A} + m_{\angle B} = 155^\circ \) needs verification:
[tex]\[ \angle A + \angle B = 25^\circ + 130^\circ = 155^\circ \quad \text{(True)} \][/tex]
- \( m_{\angle A} + m_{\angle C} = 60^\circ \) is incorrect:
[tex]\[ 25^\circ + 25^\circ = 50^\circ \][/tex]
- \( m_{\angle A}=20^\circ \) and \( m_{\angle C}=30^\circ \) is incorrect, as both \(\angle A\) and \(\angle C\) are \(25^\circ\).
Therefore, the statement that must be true is:
[tex]\[ m_{\angle A} + m_{\angle B} = 155^\circ \][/tex]
1. Identify the Given Information:
- \(\angle B = 130^\circ\)
- \(\triangle ABC\) is isosceles, implying that the two base angles \(\angle A\) and \(\angle C\) are equal.
2. Recall the Property of Triangle Angles Sum:
- The sum of the angles in any triangle is always \(180^\circ\).
3. Calculate the Base Angles:
- Since \(\triangle ABC\) is isosceles, the base angles \(\angle A\) and \(\angle C\) are equal.
- Let’s denote these angles as \(\angle A = \angle C\).
- The total sum of angles in the triangle is \(180^\circ\), so:
[tex]\[ \angle A + \angle B + \angle C = 180^\circ \][/tex]
- Substitute \(\angle B = 130^\circ\):
[tex]\[ \angle A + 130^\circ + \angle C = 180^\circ \][/tex]
Since \(\angle A = \angle C\), we can write:
[tex]\[ 2\angle A + 130^\circ = 180^\circ \][/tex]
4. Solve for \(\angle A\):
- Subtract \(130^\circ\) from both sides:
[tex]\[ 2\angle A = 50^\circ \][/tex]
- Divide both sides by 2:
[tex]\[ \angle A = \angle C = 25^\circ \][/tex]
5. Check the Provided Statements:
- \( m_{\angle A}=15^\circ \) and \( m_{\angle C}=35^\circ \) is incorrect, as both \(\angle A\) and \(\angle C\) are \(25^\circ\).
- \( m_{\angle A} + m_{\angle B} = 155^\circ \) needs verification:
[tex]\[ \angle A + \angle B = 25^\circ + 130^\circ = 155^\circ \quad \text{(True)} \][/tex]
- \( m_{\angle A} + m_{\angle C} = 60^\circ \) is incorrect:
[tex]\[ 25^\circ + 25^\circ = 50^\circ \][/tex]
- \( m_{\angle A}=20^\circ \) and \( m_{\angle C}=30^\circ \) is incorrect, as both \(\angle A\) and \(\angle C\) are \(25^\circ\).
Therefore, the statement that must be true is:
[tex]\[ m_{\angle A} + m_{\angle B} = 155^\circ \][/tex]
