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 \\).

\\( x = \\) \\( y = \\)

Explanation:

Step1: Use the property of congruent triangles

Since \(\triangle ZMK\cong\triangle APY\), corresponding angles are equal. So \(m\angle K = m\angle Y\) and \(m\angle A=m\angle Z\). Also, in a triangle, the sum of interior angles is \(180^{\circ}\). For \(\triangle ZMK\), \(m\angle Z + m\angle M+m\angle K=180^{\circ}\), and for \(\triangle APY\), \(m\angle A + m\angle P+m\angle Y = 180^{\circ}\). But because of congruence \(m\angle K=m\angle Y\) and \(m\angle A = m\angle Z\).

First, solve for \(x\):
Since \(m\angle K=m\angle Y\), we have the equation \(13x - 37=41\).
Add \(37\) to both sides: \(13x=41 + 37\).
\(13x=78\).
Divide both sides by \(13\): \(x=\frac{78}{13}=6\).

Step2: Solve for \(y\)

We know that \(m\angle Z=180-(m\angle M + m\angle K)\). Since \(m\angle M = 112^{\circ}\) and \(m\angle K = 41^{\circ}\), \(m\angle Z=180-(112 + 41)=27^{\circ}\). And since \(m\angle A=m\angle Z\), and \(m\angle A=(2y + 7)^{\circ}\), we set up the equation \(2y+7 = 27\).
Subtract \(7\) from both sides: \(2y=27 - 7\).
\(2y=20\).
Divide both sides by \(2\): \(y = 10\).

Answer:

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