Given $\tan\theta = \frac{5}{\sqrt{39}}$, set up the right triangle:

• Opposite $= 5$
• Adjacent $= \sqrt{39}$
• Hypotenuse $= \sqrt{5^2 + (\sqrt{39})^2} = \sqrt{25 + 39} = \sqrt{64} = 8$

Therefore:

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{8}$