4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{269e7aef-d7b7-4c3b-8d55-5a00696c97cc-14_888_1322_294_374}
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\caption{Figure 2}
\end{figure}
The uniform triangular lamina \(A B C D E\) is such that angle \(C E A = 90 ^ { \circ } , C E = 9 a\) and \(E A = 6 a\). The point \(D\) lies on \(C E\), with \(D E = 3 a\). The point \(B\) on \(C A\) is such that \(D B\) is parallel to \(E A\) and \(D B = 4 a\). The triangular lamina is folded along the line \(D B\) to form the folded lamina \(A B D E C F\), as shown in Figure 2.
The distance of the centre of mass of the triangular lamina from \(D C\) is \(d _ { 1 }\)
The distance of the centre of mass of the folded lamina from \(D C\) is \(d _ { 2 }\)
- Explain why \(d _ { 1 } = d _ { 2 }\)
The folded lamina is freely suspended from \(B\) and hangs in equilibrium with \(B A\) inclined at an angle \(\alpha\) to the downward vertical through \(B\).
- Find, to the nearest degree, the size of angle \(\alpha\).