Question

I need help with this practice problem, struggling It is trigonometry

Answer 1

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

`\cos \theta=(adjacent)/(hypotenuse)`

Then

`\cos \theta=\frac{2\sqrt[]{6}}{2\sqrt[]{15}}`

Now, we just need to apply the inverse property

`\theta=\arccos \mleft(\frac{\sqrt[]{6}}{\sqrt[]{15}}\mright)`

If we compute this, we obtain

Θ = 50.8°