A right triangle has a hypotenuse of length 10 units and includes a 40° angle. What are the length of the other two size

1) Let's sketch this out to better grasp it:
2) Since the sum of the interior angles is 180º, we can write out the following to find the missing angle:
[tex]\begin{gathered} 40+90+\alpha=180 \\ 130+\alpha=180 \\ \alpha=180-130 \\ \alpha=50 \end{gathered}[/tex]3) We can find the lengths by using trigonometric ratios (sine, cosine, tangent).
[tex]\begin{gathered} \sin (40)=\frac{opposite\text{ leg}}{hypotenuse}=\frac{b}{10} \\ \sin (40)=\frac{b}{10} \\ b=10\cdot\sin (40) \\ b\approx6.43 \\ \\ \sin (50)=\frac{c}{10} \\ c=10\cdot\sin (50) \\ c=7.66 \end{gathered}[/tex]