Question

10 numeric 1 point questions 9 and 10 go together y =? answer previous submit

Explanation:

Step1: Find the base angles of the isosceles triangle

Since the triangle is isosceles (two equal sides marked), the base angles are equal. Let's first find the vertex - angle related property. But for the parallel - line property, we know that when a line is parallel to one side of a triangle, it creates similar triangles. However, using the property of the isosceles triangle: the sum of angles in a triangle is \(180^{\circ}\). But for the parallel lines (the two lines with arrowheads are parallel), we use the property of similar triangles or the property of corresponding angles.

Another approach: The triangle with the \(22^{\circ}\) angle is isosceles. The base angles of the larger isosceles triangle (where the parallel lines are considered) can be found using the property of parallel lines (corresponding angles). The line segment divides the isosceles triangle into a smaller isosceles triangle (with the \(22^{\circ}\) angle) and a trapezoid - like figure. Using the property that the two parallel lines (the ones with arrowheads) and the side of the triangle as a transversal.

The angle \(x = 180-(22 + 22)=136^{\circ}\) (sum of angles in a triangle \(=180^{\circ}\) for the smaller isosceles triangle). But for \(y\), since the two lines (with arrowheads) are parallel, and using the property of the isosceles triangle, \(y = 180 - 22=158^{\circ}\) (linear - pair with the base - angle of the smaller isosceles triangle. The base - angle of the smaller isosceles triangle is \(\frac{180 - 22}{2}=79^{\circ}\), but wait, no. Wait, the correct way:

The triangle with the \(22^{\circ}\) angle: let's use the exterior - angle property. The line parallel to the base of the large isosceles triangle. The angle adjacent to \(y\) in the non - parallel part: the base - angle of the isosceles triangle (with \(22^{\circ}\) vertex angle) is \(\frac{180 - 22}{2}=79^{\circ}\). Then \(y=180 - 22=158^{\circ}\) (because of the property of the straight - line formed by the side of the triangle and the parallel line. The angle inside the triangle at the base (related to the \(22^{\circ}\) vertex) is \(79^{\circ}\), but no, wait, another way.

Since the two lines (with arrowheads) are parallel, and the side of the triangle is a transversal. The angle \(y\) and the angle adjacent to the \(22^{\circ}\) (in the sense of the larger isosceles triangle) form a linear pair. The base - angle of the isosceles triangle (with \(22^{\circ}\) vertex) is \(\frac{180 - 22}{2}=79^{\circ}\). But using the property of parallel lines (the two lines with arrowheads): the angle \(y\) and \(22^{\circ}\) (in terms of the supplementary - angle relationship for the parallel lines and the side of the triangle as a transversal). \(y = 180 - 22\)

Step2: Calculate \(y\)

Using the formula \(y=180 - 22\) (supplementary angles for the parallel lines and the side of the triangle as a transversal)
\(y = 158\)

Answer:

\(158\)