Question

1. four triangles are shown. one side of each triangle lies on a ray, and the triangles are not drawn to scale.

based on these triangles, which statement about x is true?

\bigcirc x = 160, because 180 - (130 + 30) = 20 and 180 - 20 = 160
\bigcirc x = 20, because 130 + 30 = 160 and 180 - 160 = 20
\bigcirc x = 80, because 180 - 130 = 50 and 50 + 30 = 80
\bigcirc x = 340, because 130 + 30 = 160 and 160 + 180 = 340

Explanation:

Step1: Recall triangle angle sum and linear pair

The sum of angles in a triangle is \(180^\circ\), and a linear pair (angle on a straight line) is also \(180^\circ\). For the last triangle, first find the third angle inside the triangle.

Step2: Calculate the third interior angle

The two given interior angles are \(130^\circ\) and \(30^\circ\). Let the third interior angle be \(y\). Then \(y = 180-(130 + 30)=20^\circ\) (using triangle angle - sum property: sum of interior angles of a triangle is \(180^\circ\)).

Step3: Calculate \(x\) using linear pair

Since \(x\) and \(y\) form a linear pair (they are adjacent angles on a straight line), their sum is \(180^\circ\). So \(x=180 - y\). Substituting \(y = 20^\circ\), we get \(x = 180-20 = 160^\circ\). Alternatively, we can think of \(x\) as \(180-(180-(130 + 30))=130 + 30+20?\) No, better to use the linear pair: the exterior angle \(x\) and the non - adjacent interior angles: wait, another way: the exterior angle is equal to the sum of the two non - adjacent interior angles? Wait no, in the first three triangles, let's check the pattern. In the first triangle: \(35 + 56=91\) (since \(35+56 + 91?\) Wait no, \(35 + 56=91\), and the exterior angle is \(91\). In the second triangle: \(60+60 = 120\), and the exterior angle is \(120\). In the third triangle: \(20 + 120=140\), and the exterior angle is \(140\). Oh! So the exterior angle is equal to the sum of the two non - adjacent interior angles? Wait no, in the first triangle, the two non - adjacent interior angles to the \(91^\circ\) exterior angle are \(35^\circ\) and \(56^\circ\), and \(35 + 56=91\). In the second triangle, the two non - adjacent interior angles to the \(120^\circ\) exterior angle are \(60^\circ\) and \(60^\circ\), and \(60+60 = 120\). In the third triangle, the two non - adjacent interior angles to the \(140^\circ\) exterior angle are \(20^\circ\) and \(120^\circ\), and \(20 + 120=140\). So the pattern is: the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. Wait, but in the fourth triangle, the two non - adjacent interior angles to \(x\) are \(130^\circ\) and \(30^\circ\)? Wait no, the angle adjacent to \(x\) is the third interior angle. Wait, let's re - examine. In the first triangle: interior angles are \(35^\circ\), \(56^\circ\), and \(180-(35 + 56)=89^\circ\)? Wait no, \(35+56 + 89 = 180\), and the exterior angle is \(91^\circ\), and \(180 - 89=91\). Ah, right, the exterior angle and the adjacent interior angle are supplementary (sum to \(180^\circ\)). So adjacent interior angle \(=180 - \)exterior angle. Then, using triangle angle sum, adjacent interior angle \(=180-(sum of other two interior angles)\). So \(180 - \)exterior angle \(=180-(sum of other two interior angles)\), which implies exterior angle \(=\)sum of other two interior angles. So in the fourth triangle, the two non - adjacent interior angles to \(x\) are \(130^\circ\) and \(30^\circ\)? Wait no, the angle adjacent to \(x\) is \(180 - x\). And \(130+30+(180 - x)=180\). Solving: \(130 + 30+180 - x=180\), \(240 - x = 0\), \(x = 240\)? No, that's wrong. Wait, I made a mistake earlier. Let's start over.

Correct approach: For a triangle, the sum of interior angles is \(180^\circ\), and a linear pair (exterior angle and adjacent interior angle) sums to \(180^\circ\).

In the first triangle:
Interior angles: \(35^\circ\), \(56^\circ\), and let the third be \(a\). So \(35 + 56+a=180\), \(a = 180-(35 + 56)=89^\circ\). The exterior angle is \(91^\circ\), and \(89+91 = 180\) (linear pair).

In the second triangle:
Interior angles: \(60^\circ\), \(60^\circ\), third angle \(b = 180-(60 + 60)=60^\circ\). Exterior angle is \(120^\circ\), and \(60+120 = 180\) (linear pair).

In the third triangle:
Interior angles: \(20^\circ\), \(120^\circ\), third angle \(c=180-(20 + 120)=40^\circ\). Exterior angle is \(140^\circ\), and \(40 + 140=180\) (linear pair). Wait, my earlier pattern was wrong. Wait \(35 + 56=91\) (exterior angle), \(60+60 = 120\) (exterior angle), \(20+120 = 140\) (exterior angle). Oh! Wait, \(35 + 56=91\), \(60+60 = 120\), \(20+120 = 140\). So the exterior angle is equal to the sum of the two non - adjacent interior angles. Because in the first triangle, the exterior angle \(91^\circ\) is equal to \(35^\circ+56^\circ\) (the two angles not adjacent to it). In the second triangle, exterior angle \(120^\circ\) is equal to \(60^\circ+60^\circ\) (the two angles not adjacent to it). In the third triangle, exterior angle \(140^\circ\) is equal to \(20^\circ+120^\circ\) (the two angles not adjacent to it). Ah, so the exterior angle theorem: the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles.

So in the fourth triangle, the two non - adjacent interior angles to the exterior angle \(x\) are \(130^\circ\) and \(30^\circ\)? Wait no, the angle adjacent to \(x\) is the third interior angle. Let's label the triangle: angles are \(130^\circ\), \(30^\circ\), and \(z\) (adjacent to \(x\)). Then \(z=180-(130 + 30)=20^\circ\). Then, since \(x\) and \(z\) are a linear pair, \(x + z=180\), so \(x=180 - z=180 - 20 = 160\). Which matches the first option: \(x = 160\), because \(180-(130 + 30)=20\) and \(180 - 20=160\).

Answer:

\(x = 160\), because \(180-(130 + 30)=20\) and \(180 - 20 = 160\) (the first option: \(x = 160\), because \(180-(130 + 30)=20\) and \(180 - 20 = 160\))