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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2018-11 Fallacy, 3.3 out of 5 based on 7 ratings

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