Question

4411vacr v. algebra 1 sem a cr- a
sum of interior angles of a triangle
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 = \square
(diagram: point a with line p, angles 1 and 2, triangle abc with angle at b 57°, angle at c 49°)

Explanation:

Step1: Recall triangle angle sum

The sum of interior angles in a triangle is \(180^\circ\). In \(\triangle ABC\), \(m\angle B + m\angle C + m\angle BAC = 180^\circ\). Given \(m\angle B = 57^\circ\), \(m\angle C = 49^\circ\), so \(m\angle BAC = 180^\circ - 57^\circ - 49^\circ = 74^\circ\)? Wait, no—wait, line \(p\) is parallel to \(BC\), so \(\angle 1\) and \(\angle B\) are alternate interior angles, \(\angle 2\) and \(\angle C\) are alternate interior angles? Wait, no, actually, \(\angle 1 + \angle 2 + \angle BAC = 180^\circ\) (straight line), but also, since \(p \parallel BC\), \(\angle 1 = \angle B\) (alternate interior) and \(\angle 2 = \angle C\) (alternate interior). Wait, no, let's correct: the straight line at \(A\) has \(\angle 1 + \angle 2 + \angle BAC = 180^\circ\), but in the triangle, \(\angle B + \angle C + \angle BAC = 180^\circ\). Therefore, \(\angle 1 + \angle 2 = \angle B + \angle C\)? No, wait, no—wait, actually, since \(p \parallel BC\), the transversal \(AB\) gives \(\angle 1 = \angle B\) (alternate interior angles), and transversal \(AC\) gives \(\angle 2 = \angle C\) (alternate interior angles). Wait, no, the angles at \(A\): \(\angle 1\) and \(\angle B\) are alternate interior (since \(p \parallel BC\), \(AB\) is transversal), so \(\angle 1 = \angle B = 57^\circ\). Similarly, \(\angle 2 = \angle C = 49^\circ\)? No, that can't be, because then \(\angle 1 + \angle 2 = 57 + 49 = 106\)? Wait, no, that's wrong. Wait, the sum of \(\angle 1\) and \(\angle 2\): actually, the straight line at \(A\) is \(180^\circ\), so \(\angle 1 + \angle 2 + \angle BAC = 180^\circ\). But in the triangle, \(\angle B + \angle C + \angle BAC = 180^\circ\). Therefore, \(\angle 1 + \angle 2 = \angle B + \angle C\)? No, subtract \(\angle BAC\) from both: \(\angle 1 + \angle 2 = 180^\circ - \angle BAC\), and \(\angle B + \angle C = 180^\circ - \angle BAC\). Wait, no, actually, since \(p \parallel BC\), the angles \(\angle 1\) and \(\angle B\) are equal (alternate interior), \(\angle 2\) and \(\angle C\) are equal (alternate interior). Wait, no, the direction: line \(p\) is parallel to \(BC\), so when \(AB\) is the transversal, \(\angle 1\) and \(\angle B\) are alternate interior, so \(\angle 1 = \angle B = 57^\circ\). When \(AC\) is the transversal, \(\angle 2\) and \(\angle C\) are alternate interior, so \(\angle 2 = \angle C = 49^\circ\). Then \(\angle 1 + \angle 2 = 57 + 49 = 106\)? No, that's not right. Wait, the sum of interior angles in a triangle is \(180\), so \(\angle B + \angle C + \angle BAC = 180\), so \(\angle BAC = 180 - 57 - 49 = 74\). Then, since \(\angle 1 + \angle 2 + \angle BAC = 180\) (straight line), \(\angle 1 + \angle 2 = 180 - 74 = 106\)? Wait, no, that's the same as \(57 + 49 = 106\). Wait, actually, the correct approach is: since line \(p\) is parallel to \(BC\), the consecutive interior angles or alternate? Wait, no, the key is that \(\angle 1\) and \(\angle B\) are alternate interior (so \(\angle 1 = \angle B\)), \(\angle 2\) and \(\angle C\) are alternate interior (so \(\angle 2 = \angle C\)), so \(\angle 1 + \angle 2 = \angle B + \angle C = 57 + 49 = 106\)? Wait, no, that can't be, because the sum of angles in a triangle is 180, so \(\angle B + \angle C = 57 + 49 = 106\), and \(\angle BAC = 74\), then \(\angle 1 + \angle 2 = 180 - 74 = 106\), which matches. So yes, \(\angle 1 + \angle 2 = 180 - (57 + 49)\)? Wait, no, wait: the straight line at \(A\) is \(\angle 1 + \angle 2 + \angle BAC = 180\), and in the triangle, \(\angle B + \angle C + \angle BAC = 180\), so subtracting the two equations: \((\angle 1 + \angle 2 + \angle BAC) - (\angle B + \angle C + \angle BAC) = 180 - 180\), so \(\angle 1 + \angle 2 - \angle B - \angle C = 0\), so \(\angle 1 + \angle 2 = \angle B + \angle C\)? No, that's not. Wait, no, actually, because \(p \parallel BC\), \(\angle 1 = \angle B\) (alternate interior) and \(\angle 2 = \angle C\) (alternate interior), so \(\angle 1 + \angle 2 = \angle B + \angle C = 57 + 49 = 106\). Wait, but let's check with the triangle angle sum: \(\angle B + \angle C + \angle BAC = 180\), so \(\angle BAC = 180 - 57 - 49 = 74\). Then, since \(\angle 1 + \angle 2 + \angle BAC = 180\) (straight line), \(\angle 1 + \angle 2 = 180 - 74 = 106\). Yes, that's correct. So the sum of \(\angle 1\) and \(\angle 2\) is \(106^\circ\)? Wait, no, wait, 57 + 49 is 106? 57 + 49: 50 + 40 = 90, 7 + 9 = 16, so 90 + 16 = 106. Yes. So \(m\angle 1 + m\angle 2 = 106^\circ\)? Wait, but wait, the problem is about the sum of \(\angle 1\) and \(\angle 2\). Wait, maybe I made a mistake. Wait, the triangle angle sum is 180, so \(\angle B + \angle C + \angle A = 180\), so \(\angle A = 180 - 57 - 49 = 74\). Then, since \(\angle 1 + \angle 2 + \angle A = 180\) (straight line), \(\angle 1 + \angle 2 = 180 - 74 = 106\). So the answer is 106.

Step2: Calculate the sum

Using the triangle angle sum property and the straight line at \(A\), we find that \(m\angle 1 + m\angle 2 = 180^\circ - m\angle BAC\), and \(m\angle BAC = 180^\circ - 57^\circ - 49^\circ = 74^\circ\). Thus, \(m\angle 1 + m\angle 2 = 180^\circ - 74^\circ = 106^\circ\). Alternatively, since \(p \parallel BC\), \(\angle 1 = \angle B\) and \(\angle 2 = \angle C\) (alternate interior angles), so \(m\angle 1 + m\angle 2 = m\angle B + m\angle C = 57^\circ + 49^\circ = 106^\circ\).

Answer:

\(106\)