well, we know that angle θ is in the I Quadrant, namely that sine as well as cosine are both positive and most likely 2θ is in the II Quadrant, where sine is positive, so
[tex]tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\underset{adjacent}{9}}\qquad \textit{now let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+40^2}\implies c=\sqrt{1681}\implies c=41 \\\\[-0.35em] ~\dotfill[/tex]
[tex]sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\cdot \cfrac{\stackrel{opposite}{40}}{\underset{hypotenuse}{41}}\cdot \cfrac{\stackrel{adjacent}{9}}{\underset{hypotenuse}{41}}\implies \cfrac{720}{1681}[/tex]
