Hello!
First, let's draw a triangle with the information contained in the exercise:
Notice that the exercise asks the measure of the larger acute angle (blue angle).
To find its value, we can use the formula of cosine:
[tex]\begin{gathered} \cos (\beta)=\frac{adjacent}{hypotenuse} \\ \\ \cos (\beta)=\frac{2\sqrt[]{6}}{2\sqrt[]{15}} \end{gathered}[/tex]
Let's solve this fraction by reducing and rationalizing the denominator:
[tex]\begin{gathered} \text{simplify by 2:} \\ \cos (\beta)=\frac{\cancel{2}\sqrt[]{6}}{\cancel{2}\sqrt[]{15}} \\ \\ \text{ simplify by }\sqrt[\square]{3\colon} \\ \cos (\beta)=\frac{\sqrt[]{6}}{\sqrt[]{15}}=\frac{\sqrt[]{2}}{\sqrt[]{5}} \\ \\ \cos (\beta)=\frac{\sqrt[]{2}}{\sqrt[]{5}}\frac{\cdot\sqrt[]{5}}{\cdot\sqrt[]{5}}=\frac{\sqrt[]{2\cdot5}}{\sqrt[]{5\cdot5}}=\frac{\sqrt[]{10}}{\sqrt[]{25}} \\ \\ \cos (\beta)=\frac{\sqrt[]{10}}{5} \\ \\ \cos (\beta)\cong0.632 \end{gathered}[/tex]
As we know the cosine of β is approximately 0.632, let's look at this value in the trigonometric table:
The most approximate angle is 51º.
