Q2 As shown in the figure, a triangle $ABD$ is inscribed in a semi-circle with the diameter $BC$, where
$$\mathrm{AB}=3, \quad \mathrm{BD}=5, \quad \tan\angle\mathrm{ABD}=\frac{3}{4}.$$
We are to find the lengths of the three sides $BC$, $CD$ and $DA$ of the quadrangle $ABCD$ and the area $S$ of the quadrangle $ABCD$.
First, since $\cos\angle\mathrm{ABD}=\frac{\mathbf{L}}{\mathbf{M}}$, we have $\mathrm{DA}=\sqrt{\mathbf{NO}}$.
Also, since $\sin\angle\mathrm{ABD}=\frac{\mathbf{P}}{\mathbf{Q}}$, we have $\mathrm{BC}=\frac{\square\mathbf{R}\sqrt{\mathbf{ST}}}{\mathbf{U}}$ and thus $\mathrm{CD}=\frac{\mathbf{V}}{\mathbf{W}}$. From these we obtain
$$S = \frac{\mathbf{XXY}}{\mathbf{XY}}.$$