Question

How do I find the 6 trig ratios? (With an included diagram)

Answer 1

Drawing the triangle needed

The diagram above will find hypotenuse using Pythagoras

`\begin{gathered} h^2=2^2+3^2 \\ h^2=4+9 \\ h=\sqrt[]{13} \end{gathered}`

Recall

Tan is the only positive on the 3rd quadrant others are negative

Now,

Using SOHCAHTOA

`\begin{gathered} \cos \theta=-\frac{2}{\sqrt[]{13}} \\ \sin \theta=-\frac{3}{\sqrt[]{13}} \\ \tan \theta=(3)/(2) \\ \sec \theta=-\frac{\sqrt[]{13}}{2} \\ co\sec \theta=-\frac{\sqrt[]{13}}{3} \\ \\ \text{cot}=(2)/(3) \end{gathered}`