Question

hw 4: congruent triangles
if $\triangle zmk \cong \triangle apy$, $m\angle m = 112^\circ$, $m\angle y = 41^\circ$, $m\angle k = (13x - 37)^\circ$, and $m\angle a = (2y + 7)^\circ$, find the values of $x$ and $y$.
x = \square
y = \square

Explanation:

Step1: Use Congruent Triangles Property

Since \(\triangle ZMK \cong \triangle APY\), corresponding angles are equal. So \(\angle K \cong \angle Y\) and \(\angle Z \cong \angle A\). Also, the sum of angles in a triangle is \(180^\circ\). First, find \(x\) using \(\angle K = \angle Y\) (corresponding angles of congruent triangles). So \(13x - 37 = 41\).

Step2: Solve for \(x\)

Add 37 to both sides: \(13x = 41 + 37 = 78\). Then divide by 13: \(x=\frac{78}{13}=6\).

Step3: Find \(\angle Z\) using Triangle Angle Sum

In \(\triangle ZMK\), \(\angle M = 112^\circ\), \(\angle K = 41^\circ\) (since \(\angle K=\angle Y\)). So \(\angle Z = 180 - 112 - 41 = 27^\circ\).

Step4: Solve for \(y\) using \(\angle Z = \angle A\)

Since \(\angle Z=\angle A\), \(2y + 7 = 27\). Subtract 7: \(2y = 27 - 7 = 20\). Divide by 2: \(y = 10\).

Answer:

\(x = 6\), \(y = 10\)