Question

practice c: (first taught in lesson 30) find x, y, and z.

Explanation:

Step1: Use triangle - angle sum property

The sum of angles in a triangle is $180^{\circ}$. For the right - hand triangle, if one angle is $56^{\circ}$ and the other is $2y^{\circ}$, and the third is a right - angle (not shown but assumed to be $90^{\circ}$ for a non - degenerate triangle situation), we have $56 + 2y+90 = 180$. Simplifying gives $2y=180-(56 + 90)=34$, so $y = 17$. But if we assume the two triangles are isosceles (from the arrow markings indicating equal sides), for the right - hand triangle with angles $56^{\circ},56^{\circ},68^{\circ}$ (since $180-2\times56 = 68$), and for the left - hand triangle, since the two triangles share a side and have equal - angle markings.
For the left - hand triangle, we know one angle is $60^{\circ}$. Let's use the angle - sum property of a triangle: $x+60 + z=180$. Also, from the equal - angle markings, we can assume the two triangles are congruent in a way that $z = 56$. Then $x=180-(60 + 56)=64$. And since the angle adjacent to $2y$ in the right - hand triangle and the angle in the left - hand triangle are related by the congruence of the two triangles, if we consider the non - overlapping part, and since the right - hand triangle has two equal angles of $56^{\circ}$, then $2y = 64$, so $y = 32$.

So, $x = 64$, $y = 32$, $z = 56$.

Answer:

$x = 64$, $y = 32$, $z = 56$