Question

I need help with this practice problem It’s from my trig bookI attempted this problem earlier, later, if you can, check if I am correct. I will send a pic of my work.

Answer 1

Given the right triangle:

Taking the cosine of the angle θ:

$\cos \theta=\frac{2\sqrt[]{6}}{2\sqrt[]{15}}=\frac{\sqrt[]{2}}{\sqrt[]{5}}$

Now, taking the arc cosine to find θ:

$\begin{gathered} \theta=\arccos (\sqrt[]{\frac{2}{5}}) \\ \therefore\theta\approx50.8\degree^{} \end{gathered}$