Let's start by drawing this right triangle
The larger acute angle is the one opposite to the largerst leg
We can calculate this angle, using the following
[tex]\cos \theta=\frac{adjacent}{hypotenuse}[/tex]
Then
[tex]\cos \theta=\frac{2\sqrt[]{6}}{2\sqrt[]{15}}[/tex]
Now, we just need to apply the inverse property
[tex]\theta=\arccos \mleft(\frac{\sqrt[]{6}}{\sqrt[]{15}}\mright)[/tex]
If we compute this, we obtain
Θ = 50.8°
