Question

(c) problem 16: find x, y, and z.

Explanation:

Step1: Use angle - sum property of a triangle

The sum of interior angles of a triangle is $180^{\circ}$. In the left - hand triangle, we know one angle is $62^{\circ}$ and assume the third angle is $a$. Also, in the right - hand triangle, one angle is $58^{\circ}$ and assume the third angle is $b$. Since the two triangles are adjacent and the non - labeled angles at the common vertex are vertical angles, they are equal.

Step2: Find the non - labeled angle in the right - hand triangle

For the right - hand triangle, using the angle - sum property of a triangle ($180^{\circ}$), if one angle is $58^{\circ}$ and the other non - labeled angle is $2y^{\circ}$, and the third angle is $b$. Then $b=180-(58 + 2y)$.

Step3: Find the non - labeled angle in the left - hand triangle

For the left - hand triangle, using the angle - sum property of a triangle, if one angle is $62^{\circ}$ and the other non - labeled angle is $x^{\circ}$, and the third angle is $a$. Then $a = 180-(62 + x)$.
Since $a = b$ (vertical angles), we also know that the two triangles together form a straight line at the common vertex, so the sum of the angles around the common vertex gives us another relationship. But an easier way is to note that the non - labeled angles at the common vertex are vertical angles.
We know that in the left - hand triangle, $x+62 + z=180$, and in the right - hand triangle, $58+2y + z = 180$. Also, since the two triangles are adjacent and the non - labeled angles at the common vertex are vertical angles, we can use the fact that the sum of angles in a triangle is $180^{\circ}$.
First, in the left - hand triangle, if we assume the non - labeled angle at the common vertex is $k$, then $x+62 + k=180$. In the right - hand triangle, $58 + 2y + k=180$.
We can see that $z$ and $58^{\circ}$ are corresponding angles (or vertical angles in a sense related to the parallel - like structure formed by the two triangles), so $z = 58^{\circ}$.
For the left - hand triangle, since $z = 58^{\circ}$ and one angle is $62^{\circ}$, then $x=180-(62 + 58)=60^{\circ}$.
For the right - hand triangle, since one angle is $58^{\circ}$ and $z = 58^{\circ}$, then $2y=180-(58 + 58)=64$, so $y = 32\div2=11^{\circ}$.

Answer:

$x = 60^{\circ}, y = 11^{\circ}, z = 58^{\circ}$