Question

question 13 (1 point) saved
determine the measure of n in the following diagram.
given: l₁ || l₂

Explanation:

Step1: Identify triangle angle sum

In a triangle, the sum of angles is \(180^\circ\). Also, use alternate interior angles or triangle angle properties. The two given angles in the triangle (related to the parallel lines) are \(36^\circ\) and \(29^\circ\), and angle \(n\) is an exterior angle or related to the triangle's angle sum. Wait, actually, when \(L_1 \parallel L_2\), the triangle formed has angles \(36^\circ\), \(29^\circ\), and the third angle, but angle \(n\) is equal to the sum of the two non - adjacent interior angles (exterior angle theorem).

Step2: Apply exterior angle theorem

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So, \(n = 36^\circ+ 29^\circ\)
\(n=36 + 29=65^\circ\)? Wait, no, wait. Wait, maybe I made a mistake. Wait, actually, when two lines are parallel, the triangle's angles: let's consider the triangle formed by the transversals. The angle \(n\) is vertical to the angle that is the sum of \(36^\circ\) and \(29^\circ\)? Wait, no, let's re - examine. The two parallel lines \(L_1\) and \(L_2\), with two transversals. The triangle has angles \(36^\circ\), \(29^\circ\), and the angle at \(p\) is supplementary to \(n\)? No, wait, the exterior angle theorem: the exterior angle \(n\) is equal to the sum of the two remote interior angles. So if we have a triangle where one angle is \(36^\circ\), another is \(29^\circ\), then the exterior angle \(n\) (which is adjacent to the third angle of the triangle) is equal to \(36^\circ + 29^\circ=65^\circ\)? Wait, no, wait, maybe the triangle's third angle is \(180-(36 + 29)=115^\circ\), and then \(n\) is supplementary to that? No, that can't be. Wait, no, let's look at the diagram again (mentally). The two parallel lines, \(L_1\) and \(L_2\). The angle of \(36^\circ\) on \(L_1\), \(29^\circ\) on \(L_2\). The triangle formed: the sum of the two angles ( \(36^\circ\) and \(29^\circ\)) and the angle opposite to \(n\) (vertical angle) - no, wait, the exterior angle at \(n\) is equal to the sum of the two interior angles. So \(n=36 + 29 = 65^\circ\)? Wait, no, wait, maybe I got the theorem wrong. Wait, the exterior angle theorem: in a triangle, an exterior angle is equal to the sum of the two non - adjacent interior angles. So if we have a triangle, and we extend one side, the exterior angle is equal to the sum of the two angles that are not adjacent to it. So in this case, the angle \(n\) is an exterior angle of the triangle with interior angles \(36^\circ\) and \(29^\circ\), so \(n = 36^\circ+29^\circ=65^\circ\)? Wait, no, that gives \(65^\circ\), but let's check again. Wait, maybe the triangle has angles \(36^\circ\), \(29^\circ\), and the third angle is \(180-(36 + 29)=115^\circ\), and angle \(n\) is vertical to the angle that is supplementary to \(115^\circ\)? No, that would be \(65^\circ\) as well. Wait, maybe the correct approach is: since \(L_1\parallel L_2\), the alternate interior angles, and then using the triangle angle sum. Wait, the sum of the two given angles (\(36^\circ\) and \(29^\circ\)) and the angle \(n\) - no, wait, angle \(n\) is equal to \(180-(180-(36 + 29))=36 + 29 = 65^\circ\). Yes, so \(n = 36+29 = 65\) degrees.

Answer:

\(n = 65^\circ\)