Question

I am currently working on Trigonometry, but unfortunately I am having a hard time. Here are some of my questions I've answered so far. Could someone please confirm if it is correct, and if not, please help me find the correct answer. Thank you very much.

Answer 1

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ cos(\theta)=\cfrac{\sqrt{3}}{3}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c} \\\\\\ \textit{using the pythagorean theorem}\implies c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \\\\\\ \pm\sqrt{3^2-(\sqrt{3})^2}=b\implies \pm\sqrt{6}=b$

now the root gives us the +/-, which is it? well, we know the sine is negative, so the "b" value must be negative, so it has to be -√6 = b

thus  $\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(\theta)=-\cfrac{\sqrt{6}}{3}$