# 2018-11 Fallacy

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).

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