$\begin{array}{l}a-0=4-12 \\ \Rightarrow a=-8 \\ b-1=9-0 \\ \Rightarrow b=10 \\ \text { Area }=2 \cdot \frac{1}{2}\left|\begin{array}{cc}0 & 0 \\ 12 & 1 \\ 4 & 10 \\ 0 & 0\end{array}\right|=116 \\ O C=\sqrt{145} \\ O B=\sqrt{116} \\ B C=\sqrt{(12-4)^{2}+(1-10)^{2}}=\sqrt{145} \\ \cos \theta=\frac{\sqrt{145}+\sqrt{116}-\sqrt{145}}{2}=\frac{1}{\sqrt{5}} \\ \tan \theta=2 \\ \text { Method 2 } \\ m_{O C}=\frac{1-0}{12-0}=\frac{1}{12} \\ m_{O B}=\frac{10-0}{4-0}=\frac{5}{2} \\ \tan \theta=\frac{\frac{5}{2}-\frac{1}{12}}{1+\frac{5}{2} \cdot \frac{1}{12}}=2\end{array}$
Group Event 7
Given that $O A B C$ is a parallelogram.
G7.1 Find $a$.
G7.2 Find $b$.
G7.3 Find the area of $O A B C$.
G7.4 Find $\tan \theta$.