On a math exam, there was a question that asked for the largest angle of the triangle with sidelengths \(21\), \(41\), and \(50\). A student obtained the correct answer as follows:

Let \(x\) be the largest angle. Then,
\[
\sin x = \frac{50}{41} = 1 + \frac{9}{41}.
\]
Since \( \sin 90^{\circ} = 1 \) and \( \sin 12^{\circ} 40′ 49” = 9/41 \), the angle \( x = 90^{\circ} + 12^{\circ} 40′ 49” = 102^{\circ} 40′ 49”\).

Find the triangle with the smallest area with integer sidelengths and possessing this property (that the wrong argument as above gives the correct answer).