Question

find x. 42° 44° 115° x° x =

Explanation:

Step1: Find the adjacent angle to 115°

The adjacent angle to \(115^\circ\) on a straight line is \(180^\circ - 115^\circ = 65^\circ\).

Step2: Use the exterior angle theorem or triangle angle sum

In the triangle, we know two angles: \(42^\circ\) and \(44^\circ\), and the adjacent angle we found is \(65^\circ\)? Wait, no, let's correct. Wait, the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. Wait, first, let's find the angle at the vertex of the two triangles. Wait, the angle adjacent to \(115^\circ\) is \(180 - 115=65^\circ\). Then, in the triangle with angles \(42^\circ\), \(44^\circ\), and the angle we need to relate to \(x\). Wait, actually, the exterior angle at the end (for the small triangle with angle \(x\)): the sum of the interior angles of a triangle is \(180^\circ\). Let's find the angle in the first triangle: the angle adjacent to \(65^\circ\) (wait, no, let's re - examine. The angle supplementary to \(115^\circ\) is \(65^\circ\). Then, in the triangle with angles \(42^\circ\), \(44^\circ\), and the angle opposite? Wait, no, the key is that the exterior angle of a triangle is equal to the sum of the two remote interior angles. Wait, the angle at the top: let's consider the triangle with angles \(42^\circ\), \(44^\circ\), and the angle that is supplementary to the angle related to \(x\). Wait, maybe a better way: the sum of angles around a point or using the exterior angle. Wait, the angle \(115^\circ\) is an exterior angle of a triangle. The two non - adjacent interior angles of that triangle are \(42^\circ\) and another angle. Wait, no, let's calculate the angle inside the triangle adjacent to \(115^\circ\): \(180 - 115 = 65^\circ\). Then, in the triangle with angles \(65^\circ\), \(42^\circ\), and the third angle: \(180-(65 + 42)=73^\circ\). Then, in the other triangle, we have angles \(44^\circ\), \(73^\circ\), and the angle supplementary to \(x\). Wait, no, the sum of angles in a triangle is \(180^\circ\). Wait, let's start over.

The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. The angle \(115^\circ\) is an exterior angle of a triangle. The two non - adjacent interior angles are \(42^\circ\) and the angle we can call \(y\). So \(115=42 + y\), so \(y = 115 - 42=73^\circ\). Then, in the other triangle, we have angles \(44^\circ\), \(73^\circ\), and the angle \(x\) (since they are related by a straight line or exterior angle). Wait, no, the sum of angles in a triangle is \(180^\circ\), and the exterior angle at the end is \(x\). Wait, the angle \(y = 73^\circ\), and in the triangle with angles \(44^\circ\), \(73^\circ\), and the angle supplementary to \(x\)? No, wait, the angle \(x\) and the angle formed by \(44^\circ\) and \(73^\circ\) are related. Wait, the sum of angles in a triangle is \(180^\circ\). Let's find the angle in the second triangle: \(180-(44 + 73)=63^\circ\)? No, that's not right. Wait, no, the correct approach: the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. The angle \(115^\circ\) is an exterior angle, so the two non - adjacent interior angles are \(42^\circ\) and the angle which is the sum of \(44^\circ\) and \(x\). Wait, yes! Because the exterior angle \(115^\circ\) is equal to the sum of the two non - adjacent interior angles. So \(115=42+(44 + x)\). Wait, no, that would be wrong. Wait, no, the exterior angle is equal to the sum of the two remote interior angles. Let's consider the triangle where the exterior angle is \(115^\circ\). The two remote interior angles are \(42^\circ\) and the angle that is \(44^\circ + x\)? No, maybe the other way. Let's find the angle inside the triangle: the angle adjacent to \(115^\circ\) is \(180 - 115 = 65^\circ\). Then, in the triangle with angles \(65^\circ\), \(42^\circ\), and the angle we can call \(z\), so \(z=180-(65 + 42)=73^\circ\). Then, in the triangle with angles \(z = 73^\circ\), \(44^\circ\), and the angle supplementary to \(x\), so the angle supplementary to \(x\) is \(180-(73 + 44)=63^\circ\), so \(x = 180 - 63=117^\circ\)? No, that's not right. Wait, I made a mistake. Let's use the exterior angle theorem correctly. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. The angle \(115^\circ\) is an exterior angle. The two non - adjacent interior angles are \(42^\circ\) and the angle that is \(44^\circ + x\)? No, wait, the angle \(x\) is an exterior angle of a smaller triangle. Let's look at the small triangle with angle \(x\). The two interior angles of that triangle are \(44^\circ\) and the angle that is equal to \(115^\circ-42^\circ\). Because \(115^\circ\) is an exterior angle of the triangle with angle \(42^\circ\), so the interior angle is \(115 - 42 = 73^\circ\). Then, in the small triangle, the sum of angles is \(180^\circ\), so \(x=180-(73 + 44)=63^\circ\)? Wait, no, \(73+44 = 117\), \(180 - 117 = 63\). Wait, but let's check again. The angle supplementary to \(115^\circ\) is \(65^\circ\). Then, in the triangle with angles \(65^\circ\), \(42^\circ\), the third angle is \(180-(65 + 42)=73^\circ\). Then, in the triangle with angles \(73^\circ\), \(44^\circ\), and \(x\) (since they form a triangle), \(x = 180-(73 + 44)=63\). Wait, yes, that makes sense.

Wait, no, the correct formula: exterior angle = sum of two non - adjacent interior angles. The angle \(115^\circ\) is an exterior angle. The two non - adjacent interior angles are \(42^\circ\) and \((44^\circ + x)\)? No, that would be \(115=42 + 44+x\), so \(x=115 - 42 - 44=29\)? Wait, now I'm confused. Let's draw this mentally. There is a straight line. On one side, we have a triangle with an exterior angle of \(115^\circ\). The two interior angles of that triangle are \(42^\circ\) and another angle. Then, there is another triangle attached, with an angle of \(44^\circ\) and the angle \(x\). Wait, the key is that the exterior angle of a triangle is equal to the sum of the two remote interior angles. So for the triangle where the exterior angle is \(115^\circ\), the two remote interior angles are \(42^\circ\) and \((44^\circ + x)\)? No, that can't be. Wait, let's calculate the angle inside the first triangle: \(180 - 115 = 65^\circ\). Then, in the first triangle (with angles \(42^\circ\), \(65^\circ\), and \(A\)), \(A=180-(42 + 65)=73^\circ\). Then, in the second triangle (with angles \(A = 73^\circ\), \(44^\circ\), and \(x\)), since the sum of angles in a triangle is \(180^\circ\), \(x=180-(73 + 44)=63^\circ\). Wait, but another way: the exterior angle at the end (where \(x\) is) should be equal to \(115-(42 + 44)\)? No, \(115-42 - 44 = 29\). Wait, now I see my mistake. The angle \(A = 73^\circ\) is equal to \(44^\circ+x\), because \(A\) is an exterior angle of the small triangle with angle \(x\) and \(44^\circ\). So \(73=44 + x\), so \(x=73 - 44 = 29\). Ah! That's the correct approach. Because \(A\) (which is \(73^\circ\)) is an exterior angle of the small triangle with interior angles \(44^\circ\) and \(x\). So by the exterior angle theorem, \(A=44 + x\), so \(x=A - 44\). Since \(A = 180 - 115 - 42\)? No, \(A\) is in the triangle with angles \(42^\circ\) and the angle supplementary to \(115^\circ\) (which is \(65^\circ\)). So \(A=180-(42 + 65)=73^\circ\). Then, since \(A\) is an exterior angle of the small triangle (with angles \(44^\circ\) and \(x\)), \(A = 44+x\), so \(x=73 - 44 = 29\).

Step1: Calculate the angle in the first triangle

The angle supplementary to \(115^\circ\) is \(180 - 115=65^\circ\). In the triangle with angles \(42^\circ\) and \(65^\circ\), the third angle \(A\) is \(180-(42 + 65)=73^\circ\).

Step2: Use the exterior angle theorem for the small triangle

The angle \(A = 73^\circ\) is an exterior angle of the small triangle with interior angles \(44^\circ\) and \(x\). By the exterior angle theorem, \(A=44 + x\). So \(x=A - 44=73 - 44 = 29\).

Answer:

\(29\)