Question

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\\)°

Explanation:

Step1: Recall Triangle Angle Sum

The sum of interior angles of a triangle is $180^\circ$. In $\triangle ABC$, $\angle B = 57^\circ$, $\angle C = 49^\circ$, so $\angle BAC = 180^\circ - 57^\circ - 49^\circ$.

Step2: Calculate $\angle BAC$

$\angle BAC = 180 - 57 - 49 = 74^\circ$? Wait, no—wait, line $p$ is parallel to $BC$, so alternate interior angles: $\angle 1 = \angle B$ (alternate interior, since $p \parallel BC$ and $AB$ is transversal), $\angle 2 = \angle C$ (alternate interior, $p \parallel BC$ and $AC$ is transversal). Wait, no, actually, $\angle 1$ and $\angle B$ are alternate interior, so $m\angle 1 = 57^\circ$, $m\angle 2 = 49^\circ$? Wait, no, wait the straight line at $A$: the sum of $\angle 1$, $\angle 2$, and $\angle BAC$? No, wait line $p$ and the other line (through $A$ and $C$? Wait, no, the diagram: line $p$ is parallel to $BC$, so $\angle 1$ is equal to $\angle ABC$ (alternate interior angles, transversal $AB$), and $\angle 2$ is equal to $\angle ACB$ (alternate interior angles, transversal $AC$). Then, since in triangle $ABC$, $\angle ABC + \angle ACB + \angle BAC = 180^\circ$, but $\angle 1 + \angle 2 + \angle BAC = 180^\circ$ (straight line). Wait, no, actually, the sum of $\angle 1$ and $\angle 2$: since line $p$ is parallel to $BC$, $\angle 1 = \angle B$ (alternate interior), $\angle 2 = \angle C$ (alternate interior). Then $\angle 1 + \angle 2 = \angle B + \angle C$. In triangle $ABC$, $\angle B + \angle C + \angle A = 180^\circ$, so $\angle B + \angle C = 180^\circ - \angle A$. But wait, the straight line at $A$: $\angle 1 + \angle 2 + \angle A = 180^\circ$, so $\angle 1 + \angle 2 = 180^\circ - \angle A$. But also, in triangle $ABC$, $\angle B + \angle C + \angle A = 180^\circ$, so $\angle B + \angle C = 180^\circ - \angle A$. Therefore, $\angle 1 + \angle 2 = \angle B + \angle C$. Wait, but $\angle B$ is $57^\circ$, $\angle C$ is $49^\circ$, so $57 + 49 = 106$? No, wait that can't be. Wait, no, the sum of interior angles of a triangle is $180^\circ$, so $\angle B + \angle C = 180^\circ - \angle A$. But the straight line at $A$: $\angle 1 + \angle 2 + \angle A = 180^\circ$, so $\angle 1 + \angle 2 = 180^\circ - \angle A$. Therefore, $\angle 1 + \angle 2 = \angle B + \angle C = 57 + 49 = 106$? Wait, no, wait I'm confused. Wait, let's do it properly:

1. Sum of interior angles of triangle: $m\angle B + m\angle C + m\angle BAC = 180^\circ$.
2. Line $p$ is parallel to $BC$, so by alternate interior angles: $m\angle 1 = m\angle B$ (transversal $AB$), $m\angle 2 = m\angle C$ (transversal $AC$).
3. Wait, no, that's not right. Wait, the transversal for $\angle 1$ is $AB$, so $\angle 1$ and $\angle B$ are alternate interior, so $m\angle 1 = 57^\circ$. Transversal for $\angle 2$ is $AC$, so $\angle 2$ and $\angle C$ are alternate interior, so $m\angle 2 = 49^\circ$. Then $m\angle 1 + m\angle 2 = 57 + 49 = 106$? But wait, the straight line at $A$: $\angle 1 + \angle 2 + \angle BAC = 180^\circ$, so $\angle BAC = 180 - 106 = 74^\circ$. Then in triangle $ABC$, $57 + 49 + 74 = 180$, which works. So yes, $m\angle 1 + m\angle 2 = 180 - 74 = 106$? Wait, no, wait $\angle 1 + \angle 2 = 180 - \angle BAC$, and $\angle BAC = 74$, so $180 - 74 = 106$. Alternatively, since $\angle 1 = \angle B = 57$, $\angle 2 = \angle C = 49$, so $57 + 49 = 106$. Wait, but the sum of interior angles of a triangle is $180$, so $\angle B + \angle C = 180 - \angle A$, and $\angle 1 + \angle 2 = 180 - \angle A$, so they are equal. So the sum of $\angle 1$ and $\angle 2$ is $180 - \angle A$, but also $\angle B + \angle C = 180 - \angle A$, so $\angle 1 + \angle 2 = \angle B + \angle C = 57 + 49 = 106$? Wait, no, 57 + 49 is 106? 57 + 49 = 106? 57 + 40 = 97, +9 = 106. Yes. So $m\angle 1 + m\angle 2 = 106^\circ$? Wait, but wait the problem is asking for the sum of $\angle 1$ and $\angle 2$. Wait, maybe I made a mistake. Wait, let's check again:

In triangle $ABC$, $\angle B = 57^\circ$, $\angle C = 49^\circ$, so $\angle A = 180 - 57 - 49 = 74^\circ$. Then, since line $p$ is parallel to $BC$, the angles $\angle 1$ and $\angle B$ are alternate interior, so $\angle 1 = 57^\circ$, and $\angle 2$ and $\angle C$ are alternate interior, so $\angle 2 = 49^\circ$? No, that can't be, because $\angle 1 + \angle 2 + \angle A = 180^\circ$ (straight line), so $57 + 49 + 74 = 180$, which is correct. So $\angle 1 + \angle 2 = 57 + 49 = 106$? Wait, but 57 + 49 is 106, and 180 - 74 is 106. So yes, that's correct.

Wait, but the problem is titled "Sum of Interior Angles of a Triangle", so maybe the intended approach is: since line $p$ is parallel to $BC$, $\angle 1$ is equal to $\angle B$ (alternate interior), $\angle 2$ is equal to $\angle C$ (alternate interior), so $m\angle 1 + m\angle 2 = m\angle B + m\angle C$. Then, since $m\angle B + m\angle C + m\angle A = 180^\circ$, but we don't need $\angle A$ because $\angle 1 + \angle 2 + \angle A = 180^\circ$ (straight line), so $\angle 1 + \angle 2 = 180^\circ - \angle A$, but also $\angle B + \angle C = 180^\circ - \angle A$, so they are equal. Therefore, $m\angle 1 + m\angle 2 = 57^\circ + 49^\circ = 106^\circ$? Wait, no, 57 + 49 is 106? Wait 57 + 49: 50 + 40 = 90, 7 + 9 = 16, so 90 + 16 = 106. Yes. So the sum is 106? Wait, but wait the straight line: if $\angle 1 + \angle 2 + \angle A = 180$, and $\angle A = 74$, then $\angle 1 + \angle 2 = 180 - 74 = 106$. Yes. So that's the answer.

Step1: Identify Alternate Interior Angles

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

Step2: Calculate Sum of \( \angle 1 \) and \( \angle 2 \)

Add the measures: \( m\angle 1 + m\angle 2 = 57^\circ + 49^\circ = 106^\circ \).

Answer:

\( 106 \)