Question

1. find ( mangle cxy ).

( mangle cxy=)

Explanation:

Step1: Use the SAS (Side - Angle - Side) congruence theorem

Since \(CX = CY\), \(CD = CE\) (marked as equal), and \(\angle C=\angle C\) (common angle), \(\triangle CDY\cong\triangle CEX\) by SAS. Then \(\angle CEX=\angle CDY + 62^{\circ}\).

Step2: Use the exterior - angle property of a triangle

The exterior - angle property states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. For \(\triangle CXY\), \(\angle CXY\) is an exterior angle of \(\triangle CEX\). We know that in \(\triangle CEX\), if we assume \(\angle CDY=\angle CEX - 62^{\circ}\) (from the congruence - related angle relationship). But another way:
We know that in \(\triangle CXY\), using the angle - sum property of a triangle's exterior angle. Let's first note that for \(\triangle CEX\) (where \(\triangle CDY\cong\triangle CEX\)), we can also use the formula for the exterior angle of a triangle.
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles.
We know that \(\angle CXY\) is an exterior angle of \(\triangle CEX\).
We use the formula \(m\angle CXY=\frac{1}{2}(360-(95 + 62))\) (using the property of the sum of angles around a point and the congruent - triangle - related angle relationships). But a more straightforward way:
Since \(\triangle CDY\cong\triangle CEX\) (SAS: \(CD = CE\), \(\angle C=\angle C\), \(CY = CX\)), \(\angle CEX=\angle CDY + 62^{\circ}\). Also, using the exterior - angle formula for \(\triangle CXY\) (where \(CX = CY\)).
The sum of angles in a triangle: Let's consider the fact that the exterior angle \(m\angle CXY\) and the interior angle at \(X\) in \(\triangle CXY\).
We know that \(m\angle CXY = 128.5^{\circ}\)
Another approach:
The sum of angles in a triangle: For \(\triangle CEX\) (congruent to \(\triangle CDY\)), we know that the exterior angle \(m\angle CXY\) can be found as follows.
We know that \(m\angle CXY=180-(180 - 95 - 62)\div2\) (using the property of isosceles triangles (\(CX = CY\)) and angle - sum in a triangle).
First, find the non - exterior related angle sum: \(180-(95 + 62)=23\). Then the base angles of the isosceles \(\triangle CXY\) (since \(CX = CY\)): \(\frac{23}{2}=11.5\). Then \(m\angle CXY = 180 - 11.5=128.5\)

Answer:

\(128.5^{\circ}\)