4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a3f28425-4acf-4878-b0e3-15b5bc8a92d7-12_702_1182_226_379}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
The uniform lamina \(A B C\) is an isosceles triangle with \(A B = B C , A C = 6 a\) and the distance from \(B\) to \(A C\) is \(3 a\).
The uniform lamina \(M N C\) is an isosceles triangle with \(M N = N C\) and \(M C = 3 a\). Triangles \(A B C\) and \(M N C\) are similar and are made of the same material.
The lamina \(L\) is formed by fixing triangle \(M N C\) on top of triangle \(A B C\), as shown in Figure 2.
- Show that the distance of the centre of mass of \(L\) from \(A C\) is \(\frac { 9 } { 10 } a\)
The lamina \(L\) is freely suspended from \(B\) and hangs in equilibrium.
- Find, to the nearest degree, the size of the angle between \(A B\) and the downward vertical.