The problem is to find the angle at vertex C of a triangle, given that one angle is $42^{\circ}$ and the external angle at C is $97^{\circ}$.

1. Problem Description

The problem is to find the angle at vertex C of a triangle, given that one angle is $42^{\circ}$ and the external angle at C is $97^{\circ}$.

2. Solution Steps

Let the angle at vertex A be denoted as $A$ and the angle at vertex C be denoted as $C$.

We are given that $A = 42^{\circ}$.

We are also given that the external angle at C is $97^{\circ}$.

Let the external angle at C be denoted as $C_{ext}$. Thus, $C_{ext} = 97^{\circ}$.

The external angle at any vertex of a triangle is equal to the sum of the two interior opposite angles.

Therefore, the external angle at C is equal to $A + B$, where $B$ is the angle at vertex B.

$C_{ext} = A + B$.

$97^{\circ} = 42^{\circ} + B$

$B = 97^{\circ} - 42^{\circ} = 55^{\circ}$

The sum of the angles in a triangle is $180^{\circ}$.

$A + B + C = 180^{\circ}$

$42^{\circ} + 55^{\circ} + C = 180^{\circ}$

$97^{\circ} + C = 180^{\circ}$

$C = 180^{\circ} - 97^{\circ} = 83^{\circ}$

Alternatively, the internal angle $C$ can be calculated from the external angle $C_{ext}$ as $C = 180^{\circ} - C_{ext} = 180^{\circ} - 97^{\circ} = 83^{\circ}$.

3. Final Answer

The angle at vertex C is $83^{\circ}$.