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m∠1=(4x−10)° m∠2=58° x=20 (3x−1)°=(4x−10)° ∠1=58°

Question

m∠1=(4x−10)°
m∠2=58°
x=20
(3x−1)°=(4x−10)°
∠1=58°

Explanation:

Step1: Use vertical - angle property

Vertical angles are equal. So, \((3x - 1)^{\circ}=(4x - 10)^{\circ}\)

Step2: Solve the equation for \(x\)

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Step3: Check angle - sum property (linear pair)

\(\angle1+\angle2+(3x - 1)^{\circ}=180^{\circ}\) (linear pair). Substitute \(x = 9\)
\(\angle1=(4x-10)^{\circ}=(4\times9 - 10)^{\circ}=26^{\circ}\), \((3x - 1)^{\circ}=(3\times9-1)^{\circ}=26^{\circ}\)
\(\angle1+\angle2+(3x - 1)^{\circ}=26^{\circ}+58^{\circ}+26^{\circ}=110^{\circ}
eq180^{\circ}\) (wrong).
If we use the property that \(\angle1\) and \(58^{\circ}\) are alternate interior angles (assuming parallel lines), then \(\angle1 = 58^{\circ}\)
If \(\angle1=(4x - 10)^{\circ}\), then \(4x-10 = 58\)

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If we consider \(\angle2\) and \(58^{\circ}\) (assuming \(\angle2\) and \(58^{\circ}\) are vertical angles), then \(m\angle2 = 58^{\circ}\)

Answer:

\(m\angle2 = 58^{\circ}\), \(\angle1 = 58^{\circ}\)