Question

1. if \\( \triangle z m k \cong \triangle a p y, m \angle m = 112 ^ { \circ }, m \angle y = 41 ^ { \circ }, m \angle k = ( 13 x - 37 ) ^ { \circ } \\), and \\( m \angle a = ( 2 y + 7 ) ^ { \circ } \\), find the values of \\( x \\) and \\( y \\).

Explanation:

Step1: Use the property of congruent triangles

Since \(\triangle ZMK\cong\triangle APY\), corresponding angles are equal. So \(\angle K=\angle Y\) and \(\angle M = \angle P\), \(\angle Z=\angle A\). Also, by the angle - sum property of a triangle (\(m\angle Z + m\angle M+m\angle K=180^{\circ}\), \(m\angle A + m\angle P+m\angle Y = 180^{\circ}\)).

Step2: Solve for \(x\)

Given \(m\angle K=(13x - 37)^{\circ}\) and \(m\angle Y = 41^{\circ}\). Because \(\angle K=\angle Y\) (corresponding angles of congruent triangles), we have the equation \(13x-37=41\).
Add \(37\) to both sides: \(13x=41 + 37\), so \(13x=78\).
Divide both sides by \(13\): \(x=\frac{78}{13}=6\).

Step3: Solve for \(y\)

First, find \(m\angle Z\). Using the angle - sum property in \(\triangle ZMK\): \(m\angle Z=180-(m\angle M + m\angle K)\). Substitute \(m\angle M = 112^{\circ}\) and \(m\angle K = 41^{\circ}\), then \(m\angle Z=180-(112 + 41)=27^{\circ}\).
Since \(\angle Z=\angle A\) (corresponding angles of congruent triangles) and \(m\angle A=(2y + 7)^{\circ}\), we set up the equation \(2y+7 = 27\).
Subtract \(7\) from both sides: \(2y=27 - 7\), so \(2y=20\).
Divide both sides by \(2\): \(y = 10\).

Answer:

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