Question

1. determine the value of the unknowns.

Explanation:

Step1: Find \(x\)

Vertical angles: \(x = 180^{\circ}- 115^{\circ}\)
\(x=65^{\circ}\)

Step2: Find \(y\)

Vertical angles: \(y = 180^{\circ}- 45^{\circ}\)
\(y = 135^{\circ}\)

Step3: Find \(z\)

Sum of angles in a triangle: \(74^{\circ}+55^{\circ}+(180^{\circ}-115^{\circ})=194^{\circ}\), \(180^{\circ}-(194^{\circ}-180^{\circ}) = 166^{\circ}\) (using other triangle's angle sum). Another way: Using the two - triangle relations.
For the left - hand triangle (with \(74^{\circ}\) and \(55^{\circ}\)), the third angle is \(180^{\circ}-(74^{\circ}+55^{\circ})=51^{\circ}\). For the right - hand triangle (with \(72^{\circ}\)), let's use the angle - sum property of triangles.
Since \(x = 65^{\circ}\), \(y=135^{\circ}\), and using the fact that the sum of angles in the small triangle (with \(w,x,z\)) and the relation with other angles.
\(z=74^{\circ}+55^{\circ}-(180^{\circ}-115^{\circ})=64^{\circ}\) (using exterior angle property: an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. For the large left - hand triangle, an exterior angle related to the intersection is \(180^{\circ}-115^{\circ}\), and \(z\) can be found as \(z=(74^{\circ}+55^{\circ})-(180^{\circ}-115^{\circ})\))

Step4: Find \(w\)

Sum of angles in a triangle (\(w + x+z=180^{\circ}\))
\(w=180^{\circ}-(x + z)=180^{\circ}-(65^{\circ}+64^{\circ})=51^{\circ}\)

Step5: Find \(v\)

Sum of angles in a triangle (\(72^{\circ}+v+(180^{\circ}-y)=180^{\circ}\))
Substitute \(y = 135^{\circ}\), \(72^{\circ}+v+(180^{\circ}-135^{\circ})=180^{\circ}\)
\(v=180^{\circ}-(72^{\circ}+45^{\circ})=63^{\circ}\) (also, using the property that the sum of angles in a triangle is \(180^{\circ}\). For the right - hand large triangle, if we consider the angles \(72^{\circ}\), \(v\) and the angle adjacent to \(y\) (\(180 - y\)) )

Answer:

\(x = 65^{\circ}\), \(y=135^{\circ}\), \(z = 64^{\circ}\), \(w = 51^{\circ}\), \(v=63^{\circ}\)