Problems

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Solution:

In the coordinate system where \(A\) is \((0,0)\) and \(AD\) is the \(x\)-axis and \(AB\) the \(y\)-axis and all side lengths are \(2\), we find the centre of mass of each of the sides are: \begin{align*} AB :& (0,1) \\ BC :& (1,2) \\ CD :& (2,1) \\ \\ ABCD:& \l 1, \frac{4}{3} \r \end{align*} When hung from \(A\), the angle \(AB\) makes to the vertical is \(\alpha\) and the angle \(BC\) makes to the vertical will be \(90^{\circ} + \alpha\). When hung from \(B\) the angle \(BC\) makes to the vertical will be \(\beta\). The value we are interested in therefore is \(\beta + 90^{\circ} + \alpha\) \begin{align*} && \tan \alpha &= \frac{1}{\frac{4}{3}} \\ &&& = \frac{4}{3} \\ \\ && \tan \beta &= \frac{\frac{2}{3}}{1} \\ &&&= \frac{2}{3} \\ \\ && \tan \l \beta + (90^{\circ} + \alpha) \r &= \frac{\tan \beta + \tan\l 90^{\circ} + \alpha \r}{1 - \tan \beta \tan\l 90^{\circ} + \alpha \r} \\ &&&= \frac{\frac23 + \frac43}{1- \frac23 \frac43} \\ &&&= \frac{2}{1 - \frac89} \\ &&&= 18 \end{align*}