50.8º
1) The best way to tackle this question is by sketching out this right triangle:
2) We can find the measure of the larger acute angle ( < 90º) making use of the Law of Sines. So, we can write out the following:
[tex]\begin{gathered} \frac{c}{\sin(C)}=\frac{b}{\sin (B)} \\ \frac{2\sqrt[]{6}}{\sin(C)}=\frac{2\sqrt[]{15}}{\sin (90)} \\ 2\sqrt[]{15}\cdot\sin (C)=2\sqrt[]{6}\sin (90) \\ \frac{2\sqrt[]{15}\sin(C)}{2\sqrt[]{15}}=\frac{2\sqrt[]{6}\cdot1}{2\sqrt[]{15}} \\ \sin (C)=\frac{2\sqrt[]{6}\cdot1}{2\sqrt[]{15}} \\ C=\sin ^{-1}(\frac{2\sqrt[]{6}\cdot1}{2\sqrt[]{15}}) \\ C\approx39.2^{\circ} \end{gathered}[/tex]
Note that with the help of the arcsine of these sides we can find the angle C.
Now, let's find the other angle, by using the Triangle Sum Theorem and then compare them to find which one is the larger acute angle:
[tex]\begin{gathered} 39.23+90+A=180 \\ 129.23+A=180 \\ A=180-129.23 \\ m\angle A=50.77\approx50.8^{\circ} \end{gathered}[/tex]
3) Thus the largest acute angle within that triangle is 50.8º
