[Geometry]
Problem: Let $A$ be the number you will receive from position 7 and let $B$ be the number you will receive from position 9. Let $\alpha=\sin ^{-1} A$ and let $\beta=\cos ^{-1} B$. Compute $\sin (\alpha+\beta)+\sin (\alpha-\beta)$.

Solution: The given conditions are equivalent to $\sin \alpha=A$ and $\cos \beta=B$. Using either the sumto-product or the sine of a sum/difference identities, the desired expression is equivalent to $2(\sin \alpha)(\cos \beta)=2 \cdot A \cdot B$. With $A=\frac{47}{300}$ and $B=\frac{12}{169}, 2 \cdot A \cdot B=\frac{2 \cdot 47}{25 \cdot 169}=\frac{\mathbf{9 4}}{\mathbf{4 2 2 5}}$.

Final answer: $\frac{94}{4225}$