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Question
in the diagram below, m // n and ef ⊥ gh at f. if m∠1 = 67°, what is m∠2? a 67° b 113° c 157° d 23°
Step1: Identify corresponding angles
Since \(m\parallel n\), \(\angle1\) and the angle corresponding to \(\angle1\) in the triangle are equal. Let's call this angle \(\angle3\), so \(m\angle3 = m\angle1=67^{\circ}\).
Step2: Use right - angle property
We know that \(\overline{EF}\perp\overline{GH}\), so the angle at \(F\) is \(90^{\circ}\). In the triangle with \(\angle2\), \(\angle3\) and the right - angle, we use the angle - sum property of a triangle (\(m\angle2 + m\angle3+90^{\circ}=180^{\circ}\)).
Step3: Solve for \(\angle2\)
Substitute \(m\angle3 = 67^{\circ}\) into the equation \(m\angle2 + m\angle3+90^{\circ}=180^{\circ}\). We get \(m\angle2+67^{\circ}+90^{\circ}=180^{\circ}\), then \(m\angle2=180^{\circ}-(67^{\circ} + 90^{\circ})=23^{\circ}\).
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D. \(23^{\circ}\)