smc

Let the angle be $\theta$ and the sides around it be $a$ and $b$. The area of the triangle can be written as $$A=\frac{a\cdot b\cdot \sin (\theta )}{2}$$ The doubled sides have length $2a$ and $2b$, while the angle is still $\theta$. Thus, the area is $$\frac{2a\cdot 2b\cdot \sin (\theta )}{2}$$ $$\Rrightarrow \frac{4ab\sin \theta }{2}=4A$$ $$\boxed{4}$$