Question

what are the values of x, y, and z?

Explanation:

Step1: Use angle - sum property of a triangle

In the left - hand triangle, the sum of interior angles of a triangle is $180^{\circ}$. So, for the triangle with angles $x^{\circ}$, $45^{\circ}$, and $66^{\circ}$, we have $x + 45+66 = 180$.
$x=180-(45 + 66)=180 - 111=69$.

Step2: Use angle - sum property of a triangle for the middle triangle

In the middle triangle, one angle is $30^{\circ}$ and another is $43^{\circ}$. Let's find the third angle. The sum of interior angles of a triangle is $180^{\circ}$. Let the third angle be $a$. So, $a=180-(30 + 43)=180 - 73 = 107$.

Step3: Use linear - pair and angle - sum property

We know that the angle adjacent to $119^{\circ}$ (let's call it $b$) forms a linear pair, so $b = 180 - 119=61$.
Now, for the right - hand triangle, we know one angle is $b = 61^{\circ}$ and the angle from the middle - triangle is $107^{\circ}$. Using the angle - sum property of a triangle ($y+z + b=180$) and also considering the fact that the angle at the common vertex with the middle - triangle and the right - hand triangle.
We know that $y + 43+119 = 180$ (using the fact that the angles on a straight - line and the angle - sum of a triangle). So, $y=180-(43 + 119)=180 - 162 = 18$.

Step4: Find the value of z

In the right - hand triangle, using the angle - sum property $y+z+(180 - 119)=180$. Substitute $y = 18$ into the equation: $18+z + 61=180$. Then $z=180-(18 + 61)=101$.

Answer:

$x = 69$, $y = 18$, $z = 101$