Bulletin Boards

Alumni News

Problem of the week

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).