Question

I need help with this It’s from my trigonometry prep book It asks one question within the problem

Answer 1

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:

`\begin{gathered} \cos (\beta)=(adjacent)/(hypotenuse) \\ \\ \cos (\beta)=\frac{2\sqrt[]{6}}{2\sqrt[]{15}} \end{gathered}`

Let's solve this fraction by reducing and rationalizing the denominator:

`\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}`

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.