Question

in the diagram, line p is parallel to side bc. what is the sum of the measures of ∠1 and ∠2? m∠1 + m∠2 = 106 (marked wrong)

Explanation:

Step1: Identify Alternate Interior Angles

Since line \( p \parallel BC \), \( \angle 1 \) and \( \angle B \) are alternate interior angles, so \( m\angle 1 = m\angle B = 57^\circ \). Similarly, \( \angle 2 \) and \( \angle C \) are alternate interior angles, so \( m\angle 2 = m\angle C = 49^\circ \).

Step2: Calculate the Sum

Add the measures of \( \angle 1 \) and \( \angle 2 \): \( m\angle 1 + m\angle 2 = 57^\circ + 49^\circ = 106^\circ \)? Wait, no—wait, actually, the sum of angles in a triangle is \( 180^\circ \), and the straight line at \( A \) is \( 180^\circ \). Wait, no, let's re-examine. The line \( p \) and \( BC \) are parallel, so the transversal \( AB \) gives \( \angle 1 = \angle B \) (alternate interior), and transversal \( AC \) gives \( \angle 2 = \angle C \) (alternate interior). Then, in triangle \( ABC \), \( \angle B + \angle C + \angle BAC = 180^\circ \), but \( \angle 1 + \angle 2 + \angle BAC = 180^\circ \) (straight line). Wait, no—actually, \( \angle 1 \) and \( \angle B \) are alternate interior, so \( \angle 1 = 57^\circ \), \( \angle 2 = 49^\circ \), so sum is \( 57 + 49 = 106 \)? But the triangle's angle at \( A \) would be \( 180 - 57 - 49 = 74^\circ \), and \( \angle 1 + \angle 2 + \) (angle at \( A \)) \( = 57 + 49 + 74 = 180 \), which makes sense for a straight line. Wait, but the question is the sum of \( \angle 1 \) and \( \angle 2 \), so \( 57 + 49 = 106 \)? Wait, but the initial answer was marked wrong, but maybe I misread. Wait, no—wait, the diagram: line \( p \) is parallel to \( BC \), so \( \angle 1 = \angle B = 57^\circ \) (alternate interior), \( \angle 2 = \angle C = 49^\circ \) (alternate interior). So sum is \( 57 + 49 = 106^\circ \). Wait, but maybe the problem is that the straight line is \( 180^\circ \), and the angle at \( A \) in the triangle is \( 180 - 57 - 49 = 74^\circ \), so \( \angle 1 + \angle 2 = 180 - 74 = 106^\circ \). So actually, the sum is \( 106^\circ \). Wait, but the red cross might be a mistake in the system? Or maybe I misidentified the angles. Wait, let's check again. The angle at \( B \) is \( 57^\circ \), angle at \( C \) is \( 49^\circ \). Since \( p \parallel BC \), \( \angle 1 = \angle B \) (alternate interior, because \( AB \) is transversal), \( \angle 2 = \angle C \) (alternate interior, \( AC \) is transversal). So \( \angle 1 = 57 \), \( \angle 2 = 49 \), sum is \( 57 + 49 = 106 \). So maybe the initial cross was an error, or maybe I misread the diagram. Wait, the diagram shows angle at \( B \) is \( 57^\circ \), angle at \( C \) is \( 49^\circ \), line \( p \parallel BC \). So alternate interior angles: \( \angle 1 = \angle B = 57 \), \( \angle 2 = \angle C = 49 \), so sum is \( 106 \).

Answer:

\( 106^\circ \)