∠5 and ∠6 are complementary angles. m∠5=13°

$\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}$

Step-by-step explanation:

If theta is in the third quadrant, draw the diagram to easily identify the other trigonometric relations:

Solve for the missing leg of the triangle, using the Pythagorean theorem:

$\begin{gathered} \text{ adjacent}^2+\text{ opposite}^2=\text{ hypotenuse}^2 \\ -3^2+\text{ opposite}^2=4^2 \\ \text{ opposite=}\sqrt{16-9} \\ \text{ opposite=}\sqrt{7} \end{gathered}$

Therefore, for the trigonometric relationships:

$\begin{gathered} \text{ sin\lparen}\theta)=\frac{opposite}{hypotenuse} \\ \text{ cos\lparen}\theta)=\frac{adjacent}{hypotenuse} \\ tan(\theta)=\frac{opposite}{adjacent} \\ csc(\theta)=\frac{hypotenuse}{opposite} \\ sec(\theta)=\frac{hypotenuse}{adjacent} \\ cot(\theta)=\frac{adjacent}{opposite} \end{gathered}$

Now, substitute and solve for the relations:

$\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}$