Question

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located above the 90° angle of the right triangle for reference).

Answer 1

Check the picture below.

$\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\ -------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}} \\\\\\ csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$